Question

$N _{2} O _{4}$ and $NO _{2}$ in an equilibrium mixture have partial pressure $0.70\, atm$ and $0.30\, atm$ respectively at $25^{\circ} C$ and $1\, atm$. Partial pressure of $N _{2} O _{4}$ at equilibrium at $9.0 \,atm$ and $25^{\circ} C$ is

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

Answer 1

Answer: (C)

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

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

$p_{ N _{2} O _{4}}=\frac{1-x}{1+x} \times 9=\frac{1-0.176}{1+0.176} \times 9=\frac{0.824}{1.176} \times 9=6.3 \,atm$