{"id":436615,"date":"2018-06-30T11:35:39","date_gmt":"2018-06-30T11:35:39","guid":{"rendered":"https:\/\/essaypaper.org\/?p=30551"},"modified":"2018-10-24T09:07:47","modified_gmt":"2018-10-24T09:07:47","slug":"science-chemistry-ch-chemical-equilibrium","status":"publish","type":"post","link":"https:\/\/www.benedictsol.com\/blogs\/science-chemistry-ch-chemical-equilibrium\/","title":{"rendered":"Science-Chemistry (CH- Chemical Equilibrium)"},"content":{"rendered":"

Chemistry (CH- Chemical Equilibrium)<\/h2>\n

2a-1<\/span><\/p>\n

A 0.0200-mole sample of SO3 is placed in a 1.0-liter reaction vessel and allowed to decompose until equilibrium is established according to the reaction; \u00a02 SO3(g) = 2 SO2(g) + O2(g).<\/span>
\n<\/span>At equilibrium 0.0029 mole of O2 is present. \u00a0What is the composition of the equilibrium mixture in terms of moles\/liter of each substance present?<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0[SO<\/span>3<\/span>] = 0.0142<\/span>\u00a0[SO<\/span>2<\/span>] = 0.0058<\/span>\u00a0[O<\/span>2<\/span>] = 0.0029<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2a-2<\/span><\/p>\n

A mixture of 0.296 mole NH3, 0.170 mole N2, and 0.095 mole H2, in a 1.0-liter vessel, is allowed to reach equilibrium according to the reaction; \u00a02 NH3(g) = N2(g) + 3 H2(g).<\/span>
\n<\/span>At equilibrium, it is found that 0.268 mole NH3 is present. \u00a0What is the composition of the equilibrium mixture in terms of moles\/liter of each substance present?<\/span><\/p>\n

<\/span><\/p>\n

Solution:-<\/span>\u00a0[NH3] = 0.268<\/span>\u00a0[N2] = 0.184<\/span>\u00a0[H2] = 0.137<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2a-3<\/span><\/p>\n

Write the expression for the equilibrium constant for each of the following reactions.<\/span>
\n<\/span>(a)<\/span>\u00a0SO2(g) \u00a0+ Cl2(g) \u00a0= SO2Cl2(g)<\/span>
\n<\/span>(b)<\/span>\u00a02 NO2(g) \u00a0= N2(g) + \u00a02 O2(g)<\/span>
\n<\/span>(c)<\/span>\u00a02 SO3(g) \u00a0+ CO2(g) = \u00a0CS2(g) + 4 O2(g)<\/span>
\n<\/span>(d)<\/span>\u00a04 H2(g) \u00a0+ CS2(g) \u00a0= CH4(g) + \u00a02 H2S(g)<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0(a)<\/span>\u00a0K<\/span>c<\/span><\/i> = [SO<\/span>2<\/span>Cl<\/span>2<\/span>]\/[SO<\/span>2<\/span>][Cl<\/span>2<\/span>]<\/span>\u00a0(b)<\/span>\u00a0K<\/span>2<\/span><\/i> = [N<\/span>2<\/span>][O<\/span>2<\/span>]<\/span>2<\/span>\/[NO<\/span>2<\/span>]<\/span>2<\/span>
\n<\/span>(c)<\/span>\u00a0K<\/span>c<\/span><\/i> = [CS<\/span>2<\/span>][O<\/span>2<\/span>]<\/span>4<\/span>\/[SO<\/span>3<\/span>]<\/span>2<\/span>[CO<\/span>2<\/span>]<\/span>\u00a0(d)<\/span>\u00a0K<\/span>c<\/span><\/i> = [CH<\/span>4<\/span>][H<\/span>2<\/span>S]<\/span>2<\/span>\/ [H<\/span>2<\/span>]<\/span>4<\/span>[CS<\/span>2<\/span>]<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2a-4<\/span><\/p>\n

The equilibrium constant expression for a gas-phase reaction is KC = [H2O]2[SO2]2\/[H2S]2[O2]3. Write the balanced chemical equation form which this expression is obtained.<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a02H<\/span>2<\/span>S<\/span>(g)<\/span>+3O<\/span>2(g)<\/span>=2H<\/span>2<\/span>O<\/span>(g)<\/span>+2SO<\/span>2(g)<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Solution:-<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2a-5<\/span><\/p>\n

At a particular temperature, a hypothetical chemical system has the following equilibrium molar concentrations: A = 3.00, \u00a0B = 2.00, and C = 5.00. Calculate the value of the equilibrium constant for the system if the reaction occurring were each of the following.<\/span>
\n<\/span>(a)<\/span>\u00a0A(g) \u00a0= 2 B(g) \u00a0+ C(g)<\/span>
\n<\/span>(b)<\/span>\u00a0A(g) \u00a0+ 3 B(g) \u00a0= 2 C(g) (c) 2 B(g) \u00a0= A(g) + C(s)<\/span>
\n<\/span>(d) 4 C(g) \u00a0+ B(g) = 3 A(g)<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0(a)<\/span>\u00a06.67<\/span>\u00a0(b)<\/span>\u00a01.04<\/span>\u00a0(c)<\/span>\u00a00.750<\/span>\u00a0(d)<\/span>\u00a00.0216<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2a-6<\/span><\/p>\n

A mixture of 5.000×10-3 mole H2 \u00a0and 1.000×10-2 mole of I2 is placed in a 5.000-liter container at 448oC and allowed to come to equilibrium according to the following equation. \u00a0Analysis of the equilibrium mixture shows that the concentration of HI is 1.87×10-3M. Calculate Kc at 448oC for the reaction mixture.<\/span>
\n<\/span>H<\/span>2(g)<\/span> \u00a0+ I<\/span>2(g)<\/span> \u00a0= 2HI<\/span>(g)<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0K<\/span>c<\/span><\/i> = 51<\/span><\/p>\n


\n<\/span>Solution:-<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2b-1<\/span><\/p>\n

Write the expression for the equilibrium constant for each of the following reactions.<\/span>
\n<\/span>(a)<\/span>\u00a02Pb(NO<\/span>3<\/span>)<\/span>2(s)<\/span> \u00a0= 2PbO<\/span>(s)<\/span> \u00a0+ 4NO<\/span>2(g)<\/span> \u00a0+ O<\/span>2(g)<\/span>
\n<\/span>(b)<\/span>\u00a02KClO<\/span>3(s)<\/span> \u00a0= 2KCl<\/span>(s)<\/span> \u00a0+ 3O<\/span>2(g)<\/span>
\n<\/span>(c)<\/span>\u00a02Ag<\/span>(s)<\/span> \u00a0+ Cl<\/span>2(g)<\/span> \u00a0= 2AgCl<\/span>(s)<\/span>
\n<\/span>(d)<\/span>\u00a0PCl<\/span>5(s)<\/span> \u00a0= PCl<\/span>3<\/span>(l)<\/span><\/i> \u00a0+ Cl<\/span>2(g)<\/span><\/p>\n

<\/span><\/p>\n

Ans:-(a) K<\/span>c<\/span><\/i> = [NO<\/span>2<\/span>]<\/span>4<\/span>[O<\/span>2<\/span>]<\/span>\u00a0(b)<\/span>\u00a0K<\/span>c<\/span><\/i> = [O<\/span>2<\/span>]<\/span>3<\/span>\u00a0(c)<\/span>\u00a0K<\/span>c<\/span><\/i> = 1\/[Cl<\/span>2<\/span>]<\/span>\u00a0(d)<\/span>\u00a0K<\/span>c<\/span><\/i> = [Cl<\/span>2<\/span>]<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2b-2<\/span><\/p>\n

Which one of the substances (H<\/span>2<\/span>, \u00a0H<\/span>2<\/span>O, O<\/span>2<\/span>) when added to Fe<\/span>3<\/span>O<\/span>4<\/span> in a closed container at high temperature, permits attainment of equilibrium in the reaction; \u00a03Fe<\/span>(s)<\/span>+4H<\/span>2<\/span>O<\/span>(g)<\/span>= Fe<\/span>3<\/span>O<\/span>4(s)<\/span>+4H<\/span>2(g)<\/span>?<\/span><\/p>\n

<\/span><\/p>\n

Solution:-<\/span>\u00a0Only H<\/span>2<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2b-3<\/span><\/p>\n

When the following reactions come to equilibrium, does the equilibrium mixture contain (1) mostly reactants, \u00a0(2) mostly products, or (3) appreciable concentrations of both reactants and products?<\/span>
\n<\/span>(a)<\/span>\u00a0H2(g) \u00a0+ S(s) \u00a0= H2S(g)<\/span>\u00a0KC = 7.8<\/span>105<\/span>
\n<\/span>(b)<\/span>\u00a0N2(g) \u00a0+ 2 H2(g) \u00a0= N2H4(g)<\/span>\u00a0KC = 7.4<\/span>10-26<\/span>
\n<\/span>(c)<\/span>\u00a0Cl2(g) \u00a0+ 2 NO2(g) \u00a0= 2 NO2Cl2(g)<\/span>\u00a0KC = 1.8<\/span>
\n<\/span>(d)<\/span>\u00a0H3PO4(aq) \u00a0= H+(aq) + \u00a0H2PO4-(aq)<\/span>\u00a0KC = 7.5<\/span>10-3<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0(a) <\/span>\u00a02<\/span>\u00a0(b)<\/span>\u00a01<\/span>\u00a0(c)<\/span>\u00a03<\/span>\u00a0(d)<\/span>\u00a03<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2b-4<\/span><\/p>\n

Methane (CH4) reacts with hydrogen sulfide (H2S) to yield H2 and carbon disulfide (CS2), a solvent used manufacturing rayon and cellophane; \u00a0\u00a0CH4(g) + 2H2S(g) = CS2(g) + 4 H2(g).<\/span>
\n<\/span>What is the value of Kp at 1000K if the partial pressure in an equilibrium mixture at 1000K are 0.20atm of CH4, 0.25 atm of H2S, 0.52 atm of CS2, and 0.10 atm of H2.<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0K<\/span>p<\/span><\/i> = 4.2<\/span>10<\/span>-3<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2b-5<\/span><\/p>\n

Nitric oxide reacts with oxygen to give nitrogen dioxide, an important reaction in the process for the industrial synthesis of nitric acid; \u00a02 NO(g) + O2(g) = 2 NO2(g).<\/span>
\n<\/span>(a)<\/span>\u00a0If K<\/span>C<\/span> = 6.9x<\/span>105 at 227oC, what is the value of Kp at this temperature?<\/span>
\n<\/span>(b)<\/span>\u00a0If K<\/span>p<\/span><\/i> = 1.3<\/span>10-2 at 1000K, what is the value of KC at this temperature?<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0(a) K<\/span>p<\/span><\/i> = 1.7<\/span>10<\/span>4<\/span>\u00a0(b) K<\/span>C<\/span> = 1.1<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2b-6<\/span><\/p>\n

At 700oC, KC = 20.4 for the reaction; \u00a0SO2(g) + \u00bd O2(g) = SO3(g).<\/span>
\n<\/span>At the same temperature, calculate the following equilibrium constant values.<\/span>
\n<\/span>(a)<\/span>\u00a0KC for the reaction; \u00a0SO3(g) = SO2(g) + \u00bd O2(g)?<\/span>
\n<\/span>(b)<\/span>\u00a0KC for the reaction; \u00a02 SO2(g) + O2(g) = 2 SO3(g)?<\/span>
\n<\/span>(c)<\/span>\u00a0Kp for the reaction; \u00a02 SO2(g) + O2(g) = 2 SO3(g)?<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0(a) K<\/span>C<\/span> = 0.0490<\/span>\u00a0(b)<\/span>\u00a0K<\/span>C<\/span> = 416<\/span>\u00a0(c)<\/span>\u00a0K<\/span>p<\/span><\/i> = 5.21<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2b-7<\/span><\/p>\n

The equilibrium-constant expression for a reaction is KC = [NO2]4[O2]\/[N2O5]2. \u00a0The value for the equilibrium constant at a certain temperature is 45.0. What is the value of the equilibrium constant for the reaction at the same temperature if the coefficients in the chemical reaction equation are halved and the equation is then reversed?<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0K = 0.149<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Solution:-<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2b-8<\/span><\/p>\n

Calculate the value of the equilibrium constant K3 from the values for K1 and K2.<\/span>
\n<\/span>1.<\/span>\u00a0CO(g) \u00a0+ 3 H2(g) \u00a0= CH4(g) + \u00a0H2O(g)<\/span>\u00a0K1 = 3.92 2.<\/span>\u00a0CH4(g) \u00a0+ 2 H2S(g) \u00a0= CS2(g) + 4 H2(g)<\/span>\u00a0K2 = 3.3×104 3.<\/span>\u00a0CO(g) \u00a0+ 2 H2S(g) \u00a0= CS2(g) + H2O(g) \u00a0+ H2(g)<\/span>\u00a0K3 = ?<\/span><\/p>\n

<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Ans:-<\/span>\u00a0K<\/span>3<\/span><\/i> = 1.3<\/span>10<\/span>5<\/span><\/p>\n

<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2c-1<\/span><\/p>\n

At 100<\/span>o<\/span>C the equilibrium constant for the reaction; \u00a0COCl<\/span>2<\/span>(g) \u00a0= CO(g) \u00a0+ Cl<\/span>2<\/span>(g)<\/span>
\n<\/span>has the value of KC = 2.19<\/span>10<\/span>-10<\/span>. \u00a0Are the following mixtures of COCl<\/span>2<\/span>, CO, and Cl<\/span>2<\/span> at equilibrium? \u00a0If not, indicate the direction that the reaction must proceed to achieve equilibrium.<\/span>
\n<\/span>(a)<\/span>\u00a0[COCl<\/span>2<\/span>] = 2.00<\/span>10<\/span>-3<\/span>, \u00a0[CO] = 3.31<\/span>10<\/span>-6<\/span>, \u00a0[Cl<\/span>2<\/span>] = 6.62<\/span>10<\/span>-6<\/span>
\n<\/span>(b)<\/span>\u00a0[COCl<\/span>2<\/span>] = 4.50<\/span>10<\/span>-2<\/span>, \u00a0[CO] = 1.10<\/span>10<\/span>-7<\/span>, \u00a0[Cl<\/span>2<\/span>] = 2.25<\/span>10<\/span>-6<\/span>
\n<\/span>(c)<\/span>\u00a0[COCl<\/span>2<\/span>] = 1.00<\/span>10<\/span>-2<\/span>, \u00a0[CO] = 1.48<\/span>10<\/span>-6<\/span>, \u00a0[Cl<\/span>2<\/span>] = 1.48<\/span>10<\/span>-6<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0(a)<\/span>\u00a0QC > KC \u00a0\u00a0The reaction will proceed to the left to attain equilibrium.<\/span>
\n<\/span>(b)<\/span>\u00a0QC < KC \u00a0\u00a0The reaction will proceed to the right to attain equilibrium.<\/span>
\n<\/span>(c)<\/span>\u00a0QC = KC \u00a0\u00a0The reaction mixture is at equilibrium.<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Solution:-<\/span><\/p>\n

Q = [Cl2][CO]\/[COCl2]<\/span>
\n<\/span>
\n<\/span>1. Q = 1.09 x 10^-8<\/span>
\n<\/span>since Q > Kc,<\/span>
\n<\/span>c. The reaction will proceed left to attain equilibrium.<\/span>
\n<\/span>
\n<\/span>2. Q = 5.5 x 10^-12<\/span>
\n<\/span>since Q<Kc,<\/span>
\n<\/span>b. The reaction will proceed right to attain equilibrium.<\/span>
\n<\/span>
\n<\/span>3. Q = 2.19 x 10^-10<\/span>
\n<\/span>since, Q = Kc,<\/span>
\n<\/span>a. The reaction is at equilibrium.<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2c-2<\/span><\/p>\n

A mixture of SO2, \u00a0SO3, and O2 gases is maintained in a 10.0 liter flask at a temperature at which the equilibrium constant for the reaction; \u00a02 SO2(g) + O2(g) = 2 SO3(g), is KC = 1.00<\/span>102.<\/span>
\n<\/span>(a)<\/span>\u00a0If the number of moles of SO2 and SO3 in the flask are equal, how many moles of O2 are present?<\/span>
\n<\/span>(b)<\/span>\u00a0If the number of moles of SO3 in the flask is twice the number of moles of SO2, how many moles of O2 are present?<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0(a)<\/span>\u00a00.100 mole O<\/span>2<\/span>\u00a0(b)<\/span>\u00a00.400 mole O<\/span>2<\/span><\/p>\n

<\/p>\n

<\/strong><\/p>\n

2c-3<\/span><\/p>\n

A 1.00-liter vessel contains an unknown amount of PCl<\/span>5<\/span> and 0.020 mole each of PCl<\/span>3<\/span> and Cl<\/span>2<\/span> at equilibrium at 250<\/span>o<\/span>C. \u00a0The equilibrium reaction is; \u00a0PCl<\/span>5<\/span>(g) \u00a0= PCl<\/span>3<\/span>(g) \u00a0+ Cl<\/span>2<\/span>(g).<\/span>
\n<\/span>How many moles of PCl<\/span>5<\/span> are in the vessel if KC for this reaction is 0.0415 at 250<\/span>o<\/span>C?<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0[PCl<\/span>5<\/span>] = 0.0096<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Solution:-<\/span><\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0PCl<\/span>5<\/span> (g) \u2192 PCl<\/span>3<\/span> (g) + Cl<\/span>2<\/span>(g) \u00a0\u00a0\u00a0\u00a0\u00a0<\/span>
\n<\/span>1. [PCl<\/span>3<\/span>]<\/span>equ<\/span><\/i> = 0.020 mol \/ 1.00L<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 0.020 M<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0[Cl 2] = 0.020 mol \/ 1.00L<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 0.020 M<\/span>
\n<\/span>Setting up the ICE table,<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0PCl5 (g) \u2192 PCl3(g) + Cl2(g) \u00a0\u00a0\u00a0<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Initial: <\/span>x<\/span><\/i> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00 0<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Change: – 0.020 M \u00a0\u00a0\u00a0\u00a0+ 0.020M + 0.020 M<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Equilibrium: <\/span>x<\/span><\/i> – 0.020 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.020 \u00a00.020<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Kc = [PCl3][Cl2] \/[PCl5]<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.0415 =(0.020)2 \/ (<\/span>x<\/span><\/i> – 0.020)<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>x<\/span><\/i> = 0.0296 M<\/span>
\n<\/span>[PCl 5]initially = 0.0296 M<\/span>
\n<\/span> \u00a0[PCl 5]equilibrium = <\/span>x<\/span><\/i> -0.020 = \u00a00.0296 – 0.020<\/span>
\n<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=0.0096 M<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2c-4<\/span><\/p>\n

In an equilibrium mixture of the gases N2, H2, and NH3 at 500oC, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. \u00a0The equilibrium reaction is; N2(g) + 3 H2(g) = 2 NH3(g).<\/span>
\n<\/span>What is the partial pressure of NH3 of the equilibrium mixture if Kp for this reaction is 1.45<\/span>10-5 at 500oC?<\/span><\/p>\n

<\/span><\/p>\n

Solution:-<\/span>\u00a02.24<\/span>10<\/span>-3<\/span> atm<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2c-5<\/span><\/p>\n

At 750 <\/span>o<\/span>C, K<\/span>C<\/span> = 0.771 for the reaction; \u00a0H<\/span>2(g)<\/span>+CO<\/span>2(g)<\/span>=CO<\/span>(g)<\/span>+H<\/span>2<\/span>O<\/span>(g)<\/span> If 1.00 mole of H<\/span>2<\/span> and 1.00 mole of CO<\/span>2<\/span> are placed in a 5.00 liter container and allowed to react, what will be the molar equilibrium concentrations of all species?<\/span>
\n<\/span>Ans :- <\/span>\u00a0[H<\/span>2<\/span>] = [CO<\/span>2<\/span>] = 0.106<\/span>\u00a0[CO] = [H<\/span>2<\/span>O] = 0.0937<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Solution:-<\/span><\/p>\n

Equilibrium:<\/span>\u00a0H<\/span>2(g)<\/span> \u00a0+ CO<\/span>2(g)<\/span> \u00a0= CO<\/span>(g)<\/span> \u00a0+ H<\/span>2<\/span>O<\/span>(g)<\/span>\u00a0Equilibrium Constant K<\/span>C<\/span> = 0.534<\/span><\/p>\n

Change in the number of moles of gases, <\/span>n<\/span> = <\/span>n<\/span><\/i>\u00a0products<\/span><\/i> – <\/span>n<\/span><\/i>\u00a0reactants<\/span><\/i> = 2 – 2 = 0<\/span><\/p>\n

Equilibrium constant in terms of \u00a0partial pressure, K<\/span>p<\/span><\/i> = K<\/span>C<\/span>(<\/span>RT<\/span><\/i>)<\/span>\u0394n<\/span> = K<\/span>C<\/span>(<\/span>RT<\/span><\/i>)<\/span>0<\/span> = K<\/span>C<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Equilibrium:<\/span>\u00a0H<\/span>2(g)<\/span>\u00a0 + \u00a0<\/span>\u00a0CO<\/span>2(g)<\/span> \u00a0<\/span>\u00a0= \u00a0<\/span>\u00a0CO<\/span>(g)<\/span> \u00a0<\/span>\u00a0+ \u00a0<\/span>\u00a0H<\/span>2<\/span>O<\/span>(g)<\/span><\/p>\n

Initial (<\/span>mol<\/span><\/i>):<\/span>\u00a00<\/span>\u00a00<\/span>\u00a01.00<\/span>\u00a01.00<\/span><\/p>\n

Change(<\/span>mol<\/span><\/i>):<\/span>\u00a0+<\/span>x<\/span><\/i>\u00a0+x<\/span><\/i>\u00a0-x<\/span><\/i>\u00a0-x<\/span><\/i><\/p>\n

Equilibrium (<\/span>mol<\/span><\/i>)<\/span>\u00a0x<\/span><\/i>\u00a0x<\/span><\/i>\u00a01.00<\/span>-x<\/span><\/i>\u00a01.00<\/span>-x<\/span><\/i><\/p>\n

\u00a0<\/strong><\/p>\n

Equilibrium constant, Kc = <\/span>[CO][H2O]<\/span>[CO2][H2]<\/span>\u00a0, <\/span>(1.00-x)(1.00-x)<\/span>(x)(x)<\/span>= <\/span>0.534<\/span>\u00a0x <\/span><\/i>=<\/span>\u00a00.034<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2c-6<\/span><\/p>\n

Solid NH<\/span>4<\/span>HS is introduced into an evacuated flask at 24<\/span>o<\/span>C and reacts as; NH<\/span>4<\/span>HS<\/span>(s)<\/span>=NH<\/span>3(g)<\/span>+ H<\/span>2<\/span>S<\/span>(g)<\/span>. At equilibrium the total pressure (NH<\/span>3<\/span> plus H<\/span>2<\/span>S) is 0.614 atm. \u00a0What is K<\/span>p<\/span><\/i> for this equilibrium mixture?<\/span><\/p>\n

<\/span><\/p>\n

Solution:-<\/span>\u00a0K<\/span>p<\/span><\/i> = 0.0943<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2c-7<\/span><\/p>\n

A sample of nitrosyl bromide, NOBr, decomposes according to the reaction; \u00a02 NOBr(g) = 2 NO(g) + Br2(g).<\/span>
\n<\/span>An equilibrium mixture in a 5.00-liter vessel at 100oC contains 3.22 g of NOBr, 3.08 g of NO, and 4.19 g of Br2. \u00a0(a) Calculate KC for the reaction.<\/span>
\n<\/span>(b)<\/span>\u00a0Calculate Kp for the reaction.<\/span>
\n<\/span>(c)<\/span>\u00a0What is the total pressure exerted by the mixture?<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0(a)<\/span>\u00a0K<\/span>c<\/span><\/i> = 6.44<\/span>10<\/span>-2<\/span>\u00a0(b)<\/span>\u00a0K<\/span>p<\/span><\/i> = 1.97<\/span>\u00a0(c)<\/span>\u00a0P<\/span>Total<\/span><\/i> = 0.968 atm<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

2d-1<\/span><\/p>\n

Consider the following chemical reaction at equilibrium; \u00a0CO(g) + 3 H2(g) = CH4(g) + H2O(g).<\/span>
\n<\/span>For each of the following adjustments in conditions, indicate the effect on the position of equilibrium as shifts left, shifts right, or no effect.<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Answer: -(a)<\/span>\u00a0shifts right<\/span>\u00a0(b)<\/span>\u00a0shifts left<\/span>\u00a0(c)<\/span>\u00a0shifts left<\/span>\u00a0(d)<\/span>\u00a0shifts right<\/span><\/p>\n


\n<\/span>(a)<\/span>\u00a0increase in CO concentration<\/span>
\n<\/span>(b)<\/span>\u00a0increase in CH4 concentration<\/span>
\n<\/span>(c)<\/span>\u00a0decrease in H2 concentration<\/span>
\n<\/span>(d)<\/span>\u00a0decrease in H2O concentration<\/span>
\n<\/span>
\n<\/span>2d-2<\/span><\/p>\n

Consider following chemical system at equilibrium; 2C<\/span>2<\/span>H<\/span>2(g)<\/span>+5O<\/span>2(g)<\/span>= 4CO<\/span>2(g)<\/span>+2H<\/span>2<\/span>O(g)+ heat.<\/span>
\n<\/span>For each of the following adjustments of conditions, indicate the effect on the position of equilibrium as shifts left, shifts right, or no effect.<\/span>
\n<\/span>(a)<\/span>\u00a0increasing the C2H2 concentration<\/span>
\n<\/span>(b)<\/span>\u00a0decreasing the O2 concentration<\/span>
\n<\/span>(c)<\/span>\u00a0increasing the temperature<\/span>
\n<\/span>(d)<\/span>\u00a0increasing the pressure by decreasing the volume of the container.<\/span><\/p>\n

<\/span><\/p>\n

Ans:-<\/span>\u00a0(a)<\/span>\u00a0shifts right<\/span>\u00a0(b)<\/span>\u00a0shifts left<\/span>\u00a0(c)<\/span>\u00a0shifts left<\/span>\u00a0(d)<\/span>\u00a0shifts right<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Solution:-<\/span><\/p>\n

\u00a0<\/strong><\/p>\n


\n<\/span>2d-3<\/span><\/p>\n

Consider the following chemical reaction at equilibrium; \u00a02H<\/span>2<\/span>O<\/span>(g)<\/span>+2Cl<\/span>2(g)<\/span>+heat=4HCl<\/span>(g)<\/span>+O<\/span>2(g)<\/span>.<\/span>
\n<\/span>For each of the following adjustments in conditions, indicate the effect on the position of equilibrium as shifts left, shifts right, or no effect.<\/span>
\n<\/span>(a)<\/span>\u00a0heating the equilibrium mixture<\/span>
\n<\/span>(b)<\/span>\u00a0increasing the size of the mixture\u2019s container<\/span>
\n<\/span>(c)<\/span>\u00a0increasing the pressure on the equilibrium mixture by adding unreactive He gas<\/span>
\n<\/span>(d)<\/span>\u00a0adding a catalyst to the equilibrium mixture<\/span><\/p>\n

<\/span><\/p>\n

Ans:-(a)<\/span>\u00a0shifts right<\/span>\u00a0(b)<\/span>\u00a0shifts right<\/span>\u00a0(c)<\/span>\u00a0no effect<\/span>\u00a0(d)<\/span>\u00a0no effect<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Solution:-<\/span><\/p>\n


\n<\/span>2d-4<\/span><\/p>\n

Consider the following chemical reaction at equilibrium; N2(g) \u00a0+ 3 H2(g) = 2 NH3(g).<\/span>
\n<\/span>KC = 5.81 at 350oC. \u00a0For an equilibrium mixture composition [N2] = 0.885, \u00a0[H2] = 0.665, and [NH3] = 1.230, indicate how each of the following adjustments in conditions will affect the position of the equilibrium using shifts left, shifts right, or no effect.<\/span>
\n<\/span>(a)<\/span>\u00a0[N2] is increased by 0.200<\/span>
\n<\/span>(b)<\/span>\u00a0[NH3] is decreased by 0.200<\/span>
\n<\/span>(c)<\/span>\u00a0[N2] is increased by 0.300 and [H2] is increased by 0.200<\/span>
\n<\/span>(d)<\/span>\u00a0[N2] is decreased by 0.300 and [H2] is increased by 0.300<\/span><\/p>\n

\u00a0<\/strong><\/p>\n

Ans:- (a) shifts right<\/span>\u00a0(b) shifts right<\/span>\u00a0(c). \u00a0shifts right (d)<\/span>\u00a0QC < KC; shifts right<\/span>
\n<\/span>
\n<\/span>2d-5<\/span><\/p>\n

In a 10.0-liter mixture of H2, I2, and HI at equilibrium at 425oC there are 0.100 mole of H2, 0.100 mole of I2, and 0.740 mole of HI. \u00a0The equilibrium reaction is; H2(g) + I2(g) = 2 HI(g).<\/span>
\n<\/span>
\n<\/span>If 0.50 mole of HI is added to this system, what will be the molar concentrations of all the species once equilibrium is reestablished?<\/span><\/p>\n

<\/span><\/p>\n

Solution:-<\/span>\u00a0[H<\/span>2<\/span>] = [I<\/span>2<\/span>] = 0.0153<\/span>\u00a0[HI] = 0.1134<\/span><\/p>\n


\n<\/span>2d-6<\/span><\/p>\n

An equilibrium mixture, \u00a0CO2(g) + H2(g) = CO(g) \u00a0+ H2O(g)
\nwas found to contain 0.20 mole of H2, 0.80 mole of CO2, 0.10 mole of CO, and 0.40 mole of H2O in a 1.0-liter vessel. \u00a0How many moles of CO2 would have to be added to the equilibrium mixture at constant temperature and volume to increase the amount of CO to 0.20 mole?<\/p>\n

Ans:-\u00a03.3 moles\/liter<\/p>\n","protected":false},"excerpt":{"rendered":"

Chemistry (CH- Chemical Equilibrium) 2a-1 A 0.0200-mole sample of SO3 is placed in a 1.0-liter reaction vessel and allowed to decompose until equilibrium is established according to the reaction; \u00a02 SO3(g) = 2 SO2(g) + O2(g). At equilibrium 0.0029 mole Read More …<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[15],"tags":[],"class_list":["post-436615","post","type-post","status-publish","format-standard","hentry","category-essay-paper-writing"],"_links":{"self":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/436615","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/comments?post=436615"}],"version-history":[{"count":0,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/436615\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/media?parent=436615"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/categories?post=436615"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/tags?post=436615"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}