Question

Ammonium carbonate dissociates as:
$N{H}_{2}COON{H}_{4\left(S\right))}\leftrightharpoons 2N{H}_{3\left(g\right)}+C{O}_{2\left(g\right)}$
In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that the partial pressure of $N{H}_3$ now equals to the original total pressure. The ratio of total pressure now to the original pressure is:

• A. $\frac{27}{31}$
• B. $\frac{31}{27}$
• C. $\frac{4}{9}$
• D. $\frac{5}{9}$

Answer 1

Answer: (B)

$\frac{31}{27}$

$N{H}_{2}COON{H}_{4\left(S\right)}\rightleftharpoons 2N{H}_3+C{o}_2$

$Initially:2P_o{P}_o$

$K_P={\left[P_{N{x}_3}\right]}^2\left[P_{C{O}_2}\right]=4P_{\circ }^3\dots \dots \left(1\right)$

$P_{total}initial=2P_{\circ }+P_{\circ }=3P_{\circ }$

$N{H}_{2}COON{H}_4\rightleftharpoons 2N{H}_3+C{O}_2$

$3P_{\circ }{P}^{\prime }$

Finally:

$K_P=(3P_{\circ }{)}^2{P}^{\prime }$

$\therefore 4P_{\circ }^3=9P_{\circ }^2.P^{\prime }\dots \dots$ [From $\left(1\right)$]

$\therefore P^{\prime }=\frac{4}{9}{P}_{\circ }$

$\left(P_{total}\right)new=3P_{\circ }+P^{\prime }=\frac{4P_{\circ }}{9}+3P_{\circ }=\frac{31P_{\circ }}{9}$

$\therefore$ Ratio of new total pressure to original total pressure,

$=\frac{\frac{31P_{\circ }}{9}}{3P_{\circ }}=\frac{31}{27}$