Question

An industrial chemist introduces 2.5 atm H2 and 2.5 atm CO2 into a 1.00-L container at 25.0°C and then raises the temperature to 700.0°C, at which Keq = 0.534: H2(g) + CO2(g) ⇔ H2O(g) + CO(g) How many grams of H2 are present after equilibrium is established?

Answer 1

Assuming ideal gas behavior partial pressure and amount of substance can be given as---
P.X∙V = n.X∙R∙T

So ,[X] = n.X/V = P.X/(R∙T)

Since both H₂ and CO₂ are introduced at the same initial partial pressure they have the same initial concentration given by---

2.5 atm = 2.5 x 101325 Pa, R = 8.3145 Pa∙m³∙K^-1∙mol^-1 , T = ( 25 +273) K = 298 K

C₀ = 2.5 * 101325 Pa / ( 8.3145 Pa∙m³∙K^-1∙mol^-1 * 298 K)

=> C₀= 253312.5 / 2477.721 mol∙m^-3

=> C₀ = 102.2 mol∙m^-3

=> C₀ = 0.1022 M

The reaction is--
H₂ (g) + CO₂ (g) <-----> H₂O(g) + CO(g)

So the concentrations in equilibrium state satisfy the relation

K_eq = ( [H₂O]∙[CO] ) / ( [H₂]∙[CO₂] )

ICE table can be written as---
_____ [H₂]______[CO₂]____ [H₂O]_____[CO]

I_____ C₀.............. C₀............. 0................ 0

C_____ - x............. - x........... + x............. + x

E_____C₀ - x......... C₀ - x........... x............... x

Now, K_eq = ( x∙ *x ) / ( (C₀ - x)*(C₀ - x) )

=> K_eq = x² /(C₀ - x)²

=>K_eq = x /(C₀ - x)

=> (C₀ - x) *K_eq = x

=> C₀*K_eq - x *K_eq = x

=> x ( 1 + K_eq) = C₀*K_eq

=> x = C₀∙K_eq / (1 + K_eq)

Now,
The concentration of hydrogen in equilibrium state is:

[H₂] = C₀ - x
=> [H₂] = C₀ - {C₀∙K_eq /(1 + K_eq) }

= C₀ ∙/(1 + K_eq)

= 0.1022 M / (1 + 0.534)

= 0.1022 M / 1 + 0.73

= 0.1022 M / 1.73

= 0 0591 M

So the amount of hydrogen in the container is

nH₂ = [H₂]∙V = 0.0591 mol/L x 1.00 L = 0.0591 mol

and the mass of hydrogen in the equilibrium mixture will be---

mH₂ (mass) =moles x molar mass = nH₂ x MH₂ = 0.0591 mol x 2.0 g/mol = 0.1182 g