Question

A 1.604-g sample of methane (CH₄) gas and 6.400 g of oxygen gas is sealed in a 2.50-L vessel at 411⁰C and is allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is 0.326 atm, and water vapor pressure is 4.45 atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

Short answer

The pressure of carbon monoxide at equilibrium is: Pco = 0.58 atm

The pressure of CO₂ at equilibrium is: Pco₂ =1.65

Step-by-step solution

First, we have to find out the initial pressure of oxygen by using the ideal equation From the given value, the volume, number of moles, gas constant, and temperature:

$\mathbf{PV}\mathbf{=}\mathbf{nRT}\mathbf{}\mathbf{}\mathbf{}$

${\mathrm{P}}_0\left({\mathrm{forO}}_2\right)={\mathrm{n}}_{{\mathrm{O}}_2}\frac{\mathrm{RT}}{\mathrm{V}}$

On putting values in the equation

${\mathrm{P}}_0=\frac{(6.400\mathrm{g}\times 0.08206\times 684\mathrm{K})}{(32\mathrm{g}/\mathrm{mol}\times 2.50\mathrm{L})}$

We get P₀ =1.46 of oxygen

Chemical equation 1: ${\mathbf{CH}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Change occurs during reaction:-x+-2x→+x+2x

Chemical equation 2: ${\mathbf{CH}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\frac{\mathbf{3}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Change -y+ y + y + 2y

Amount of O₂ reacted = 4.49 atm – 0.326 atm = 4.16 atm of O₂

Now, when we relate the equation and the value of O2 and H2O, i.e., water vapor

We get $-2\mathrm{x}-\frac{3}{2}\mathrm{y}=4.16\mathrm{atm}{\text{O}}_2$and$2\mathrm{x}+2\mathrm{y}=4.45{\mathrm{atmH}}_{\text{2}}\text{O}$

On solving using simultaneous equations:

$2\mathrm{x}+2\mathrm{y}=4.45\mathrm{atm}$

$-2\mathrm{x}-\frac{3}{2}\mathrm{y}=4.16\mathrm{atm}$

(0.58) y =0.29,

Pco = y = 0.58 atm

2x + 2(0.58) = 4.45

$\mathrm{x}=\frac{4.45-1.16}{2}=1.645$

Hence, the pressure of carbon monoxide at equilibrium is: Pco = 0.58 atm

the pressure of CO₂at equilibrium is: Pco₂ = 1.645