Question

$N_2{O}_4$ and $N{O}_2$ in an equilibrium mixture have partial pressure $0.70atm$ and $0.30atm$ respectively at $25^{\circ }C$ and $1atm$. Partial pressure of $N_2{O}_4$ at equilibrium at $9.0atm$ and $25^{\circ }C$ is

• A. $0.77atm$
• B. $1.01atm$
• C. $6.3atm$
• D. $8.69atm$

Answer 1

Answer: (C)

Total gaseous moles at equilibrium $=1-x+2x=1+x$

Now, $p_{N_2{O}_4}=0.7=\frac{1-x}{1+x}\times 1$

From above calculation, $x=\frac{0.3}{1.7}=0.176$

Now, $p_{N_2{O}_4}$ at equilibrium = mole fraction $\times$ total pressure

$p_{N_2{O}_4}=\frac{1-x}{1+x}\times 9=\frac{1-0.176}{1+0.176}\times 9=\frac{0.824}{1.176}\times 9=6.3atm$