i need help with this problem. If you can show all work i would really appreciat

Question

i need help with this problem. If you can show all work i would really appreciate it.

A mixture of 0.200 mol of CO_2, 0.100 mol of H_2 and 0.160 mol of H_2O is place in a 2.00 L vessel. The following equilibrium is established. CO_2 (g) + H_2 (g) rightarrow CO (g) + H_2O(g) At equilibrium, [H_2O] = 0.0956 M. Calculate the equilibrium concentrations of CO_2, H_2 and CO. Calculate K_c for this reaction. Given the K_c for the reaction determined in b), what is the K_c of the following reaction: CO_2(g)+3 H_2(g) rightarrow 3 CO(g) + 3 H_2 O(g)

Explanation / Answer

M = mol/V

n = 0.2/2 = 0.1 M CO2

n = 0.1/2 = 0.05 M H2

n = 0.16/2 = 0.08 M H2O

[H2O] eq = 0.0956

then

state all concnetrations in equilibrium (stoichiometrically)

[CO2] = 0.1 -x

[CO] = 0 + x

[H2] = 0.05 - x

[H2O] = 0.08 + x

and

[H2O] = 0.08 + x = 0.0956

x = 0.0956-0.08 =0.0156

[CO2] = 0.1 -0.0156 = 0.0844

[CO] = 0 + 0.0156 = 0.0156

[H2] = 0.05 - 0.0156 = 0.0344

[H2O] = 0.08 + 0.0156 = 0.0956

b)

Kc reaction

Kc = [CO][H2O]/[CO2][H2]

then

Kc = (0.0156 *0.0956)/(0.0844*0.0344) = 0.51366

Kc = 0.51366

c)

For

3CO2 + 3H2 = 3CO + 3H2O

the Kc is affected by 3

i.e.

Kc new = 3*CO * 3H2O / (3*CO2*3H3)

Kc is affected by ^3

then

Kc new = 0.51366^3 = 0.13552

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