Answer:
a) Kc = 0.578
b) [CO2] = 0.4764 M
[H2]= 0.0164 M
[CO]= 0.0756 M
[H2O]=0.0596 M
Explanation:
Step 1: Data given
Temperature = 686 °C
The equilibrium concentrations of the reacting species are:
[CO] = 0.0520 M
[H2] = 0.0400 M
[CO2] = 0.0810 M
[H2O] = 0.0360 M
Step 2: The balanced equation
CO2(g) + H2(g) ⇌ CO(g) + H2O(g)
For 1 mol CO2 we need 1 mol H2 to produce 1 mol CO and 1 mol H2O
Step 3: Calculate Kc
Kc = [CO][H2O] / [CO2][H2]
Kc = (0.0520 * 0.0360) / (0.0810 * 0.0400)
Kc = 0.578
b) If we add CO2 to increase its concentration to 0.500 mol / L, what will the concentrations of all the gases be when equilibrium is reestablished?
adding CO2 the equilibrium will shift on the right:
Concentrations at the equilibrium will be:
[CO2] = 0.500 -x
[H2]= 0.0400 -x
[CO]= 0.0520+x
[H2O]= 0.0360+x
0.578 = ( 0.0520+x)( 0.0360+x)/ ( 0.500-x)( 0.0400-x)
0.578 = ( 0.0520+x)( 0.0360+x)/ (0.02 -0.500x -0.0400x +x²)
0.01156 - 0.26588 + 0.578x² = 0.001872 + 0.0880x + x²
x = 0.0236
[CO2] = 0.500 - 0.0236 = 0.4764 M
[H2]= 0.0400 - 0.0236 = 0.0164 M
[CO]= 0.0520+0.0236 = 0.0756 M
[H2O]= 0.0360+ 0.0236 = 0.0596 M
