A little Quiz on Chemical Equilibrium

True or False

Consider the chemical equilibrium of the following reaction:

2 SO2 (g) + O2 (g) <--> 2 SO3 (g) + heat

Consider the Equilibrium:

N2 (g) + 3H2 (g) <--> 2NH3 (g), Kp = 1 x 10-4

According to Le Chatelier's Principle:



Consider the reaction:

4HCl(aq) + MnO2 (s) <--> Cl2 (g) + 2H2O(l) + MnCl2 (aq)

The equilibrium is displaced to the left if:

Consider the reaction:

CH4 (g) + 4Cl2 (g) <--> CCl4 (liq) + 4 HCl(g) H0 = -398 kJ/mol

The equilibrium is displaced to the right if:


Long Answer



What substances present in the chemical reaction would NOT be included in the equilibrium expression?


answer:


For

CO2 (g) + H2 (g) <--> CO(g) + H2O(g)

If there are 1.43 moles each of CO and H2O, 0.572 moles H2 and 4.572 moles CO2, in a 4.0 L container at equilibrium, what is KC ?


Answer: 0.782

Consider the decomposition

2 SO3 (g) <-> 2 SO2 (g)+ O2(g).

At equilibrium, there are 0.090 mol SO2 , 0.110 mol O2 , 0.100 mol SO3 in a 25.0-L container. What is the value of KC ?

Answer: 3.6 x 10-3



For the reaction

3Fe(s) + 4H2O(g) <--> Fe3O4 (s) + 4H2 (g)
Write the expression for Kp




ANSWER: Kp = {P(H2)}4 /{P(H2O)}4

Write the expression for KC in terms of Kp




ANSWER: KC = Kp

What is the effect of adding Fe(s)?





ANSWER: no change

What is the effect of removing H2 ?





ANSWER: shift to right

A mixture of 0.75 mol of N2 and 1.20 mol of H2 are placed in a 3.0 liter container. When the reaction

N2 (g) + 3H2 (g) <--> 2NH3 (g)

reaches equilibrium, [H2] = 0.100 M. What is the value of [N2] and [NH3] at equilibrium?

A mixture of 2.5 moles H2O and 100 g of C are placed in a 50-L container. At equilibrium for the reaction

C(s) + H2O(g) <--> CO(g) + H2(g),

[H2] = 0.040 M. What is the value of [H2O] at equilibrium?


Answer: [H2O] = 0.010 M

The equilibrium constant, Kp , equals 1.78 at 250oC for the decomposition reaction:

PCl5 (g) <--> PCl3 (g) + Cl2 (g)
Calculate the percentage of PCl5 that dissociates if 0.05 mole of PCl5 is placed in a closed vessel (constant volume) at 250oC and at 2.00 atm pressure.




Solution:

V = (0.05 mol)(0.082 l atm/mol K)(523 K)/(2.00 atm) = 1.072 liters

Kc = 1.78/((0.082 l atm/mol K)(523K)) = 0.0414

Kc = (x/1.072)2/((0.05 - x)/1.072))

0.0445 = x2/(0.05 - x)

x = 0.03 (successive approx)

%dissociated = 0.03/0.05 x 100 = 60%


Consider the endothermic reaction:

N2 (g) + O2 (g) <--> 2NO(g), H0 = 192.5 kJ/mol

At 2000 K the equilibrium constant is 5.0 x 10-4. At 2500 K the value of the equilibrium constant is:




solution:

ln(K2/K1) = -H/R (1/T2 - 1/T1)
K2 = 5.1 x 10-3


PJ Brucat // University of Florida