Question

7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s) ⇀↽ 2 NH3(g) + CO2(g) The total pressure in the closed container under these condition is found to be 0.843 atm. Calculate a value for the equilibrium constant, Kp. A) 0.00701 B) 0.0888 C) 0.222 D) 0.599

Answer 1

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both $NH_{3}$ and $CO_{2}$ are gaseous. Hence equilibrium constant depends upon partial pressures of $NH_{3}$ and $CO_{2}$.

Initially no $NH_{3}$ and $CO_{2}$ were present.

Hence mole fraction of $NH_{3}$ and $CO_{2}$ at equilibrium can be calculated from coefficient of $NH_{3}$ and $CO_{2}$ in balanced equation.

Mole fraction of $NH_{3}$ = (number of moles of $NH_{3}$)/(total number of moles of $NH_{3}$ and $CO_{2}$) = $\frac{2moles}{(2+1)moles}=\frac{2}{3}$

Mole fraction of $CO_{2}$ = (number of moles of $CO_{2}$)/(total number of moles of $NH_{3}$ and $CO_{2}$) = $\frac{1moles}{(2+1)moles}=\frac{1}{3}$

Let's assume both $CO_{2}$ and $NH_{3}$ behaves ideally.

Therefore partial pressure of $NH_{3}$, $P_{NH_{3}}= x_{NH_{3}}.P_{total}$ and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, $P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm$

$P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm$

So, $K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888$