Question

In the reaction:
$\mathrm{NH}_{4} \mathrm{COONH}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})$
Equilibrium pressure for this reaction is \[3~\text{atm}\]. The value of \[^{K_{p}}\] will be:

Answer 1

Hint: We know that the number of moles of a gas varies directly with the partial pressure at constant temperature and volume. We can use this to calculate the partial pressures of each component in a reaction.

Complete step by step answer:
The given chemical equation is shown below.
$\mathrm{NH}_{4} \mathrm{COONH}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})$
We can assume that at equilibrium, the partial pressures of two moles of \[\text{NH}_3\left ( \text {g} \right )\] and one mole of \[\text{CO}_2\left ( \text {g} \right )\] be \[2p\] and \[p\] respectively.
We know from the question that the total pressure of the reaction at equilibrium is \[3~\text{atm}\].
Therefore, we can calculate the value of p as follows
$\Rightarrow$ 2p+p=3atm
$\Rightarrow$ 3p=3atm
$\Rightarrow$ p=1atm
We know that the expression for \[K_p\] for the given reaction is written as follows.
$K_p\;=(p_{NH_3})^2\times(p_{CO_2})$

When we substitute the value of partial pressures of $NH_3$(g) and $CO_2$(g) in the above expression, we can calculate the value of p.
$
K_{p}={(2 p)^{2}}(p) \\
= 4 p^{3}$ $\:\:\:\:$ Since p = 1
= 4

Hence, we can say that the value of \[{K_{p}}\] is 4.

Note: We know that in the expression of \[{K_{p}}\] the partial pressure of gaseous products and gaseous reactants are written each raised to the power of their stoichiometric coefficients. It is also known that the pressure of pure solids is taken to be negligible and is included in the value of \[{K_{p}}\].