Be sure to answer all parts. Consider the following equilibrium process at 686°C:CO2(g) + H2(g) ⇌ CO(g) + H2O(g)The equilibrium concentrations of the reacting species are [CO] = 0.0520 M, [H2] = 0.0400 M, [CO2] = 0.0810 M, and [H2O] = 0.0360 M.(a) Calculate Kc for the reaction at 686°C.(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?

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

i think   the answer is c hope i'm right

1. a)c-14 has two more protons

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3. c) iron and nickel.