Calculating pqrtial pressure at equilibrium

Question

C(s)+ Co2(g)2CO (g).at equilibrium,25% of the co2 got converted into CO. if the equilibrium pressure was 12 atm. The partial pressure of co2 at equilibrium was.   

1)0.25 atm

2)7.2 atm

3)2.4 atm

4)9 atm

Answer 1

Correct option is 2) 7.2 atm

\(C_{(s)} + \underset{x - \frac x4}{CO_{2(g)}} ⇔ \underset{\frac x2}{2CO_{(g)}}\)

Let us take initial no. of moles of CO₂ as x

\(CO_2 \to \frac{3x }4\)

\(CO \to \frac x2\)

\(\frac{3x}{4} + \frac x2\) → 12 atm

\(\frac{5x}{4} \) → 12 atm

\(x \to \frac{12 \times 4}{5} = \frac{48}5\)

= 9.6 atm

x moles → 9.6 atm

No. of moles of CO₂ at equilibrium = \(\frac{3x}{4} \)

\(\frac{3x}{4} = \frac 34 \times 9.6\)

= 7.2 atm

Partial pressure of CO₂ at equilibrium = 7.2 atm