Mixture of $3.00$volumes of${\text{H}}_2$ and$1.00$ volume of${\text{N}}_2$ reacts atrole="math" localid="1657001296390" ${344}^{°}\text{C}$ to form ammonia. The equilibrium mixture at 110atm contains $41.49\%{\text{NH}}_3$by volume. Calculate $K_{\text{p}}$for the reaction, assuming that the gases behave ideally.

A substance's volume is the amount of space it takes up, whereas its mass is the amount of stuff it contains.

In this problem, we are tasked to solve for $K_p$of the reaction. The reaction is shown below based on the description in the problem.

${\text{H}}_2\left(g\right)+{\text{N}}_2\left(g\right)\rightleftharpoons {\text{NH}}_3\left(g\right)$

at $T={344}^{°}\text{C}$.

First, balance the reaction equation. We can start by balancing the nitrogen.

${\text{H}}_2\left(g\right)+{\text{N}}_2\left(g\right)\rightleftharpoons 2{\text{NH}}_3\left(g\right)$

Then, balance the hydrogen.

.$3{\text{H}}_2\left(g\right)+{\text{N}}_2\left(g\right)\rightleftharpoons 2{\text{NH}}_3\left(g\right)$

Next, solve for the total volume of the system.

$V_{\text{total}}={\text{volumeofH}}_2+{\text{volumeofN}}_2$

$\begin{array}{l}{V}_{\text{total}}=3.00+1.00\\ V_{\text{total}}=4.00\end{array}$

Therefore the total volume is .$4.00$

Solve for the mole ratio of each reactant.

$\begin{array}{l}{X}_{{\text{II}}_2}=\frac{3.00\text{volumes}}{4.00\text{volumes}}=0.75\\ X_{{\text{N}}_2}=\frac{1.00\text{volumes}}{4.00\text{volumes}}=0.25\end{array}$

Solve for the amount of ammonia at equilibrium by multiplying the total pressure at equilibrium by the percentage of ammonia.

$P_{{\text{NH}}_{2,0,9}}=110\text{atm}\times 0.4149=45.6\text{atm}$

The partial pressure of each reactant is the product of the mole ratio and the total pressure. $P=X\cdot P$total. Write the reaction table using the given values.

Solve for from the column of the reaction table for ammonia.

$P_{{\text{NH}}_{3,\text{eq}}}=P_{{\text{NH}}_3}+2x$

$2x=P_{{\text{NH}}_{\text{seq}}}-P_{{\text{NH}}_3}$

$x=\frac{P_{{\text{NH}}_{3,\text{mq}}}-P_{{\text{NH}}_3}}{2}$

$=\frac{45.6\text{atm}-0}{2}$

$x=22.8$atm

We know that the $P_{\text{total,eq}}=110$, equate the equilibrium formula to the total pressure and substitute the calculated value for x.

$P_{\text{total,eq}}=P_{{\text{H}}_{2,\text{eq}}}+P_{{\text{N}}_{2,\text{eq}}}+P_{{\text{N}}_{\text{a},\text{eq}}}$

$110\text{atm}=\left(0.75\cdot P_{\text{total}}-3x\right)+\left(0.25\cdot P_{\text{total}}-x\right)+45.6\text{atm}$

$110\text{atm}=0.75\cdot P_{\text{total}}-3(22.8\text{atm})+0.25\cdot P_{\text{total}}-(22.8\text{atm})+45.6\text{atm}$

$110\text{atm}=1.0P_{\text{total}}-45.6\text{atm}$

$P_{\text{total}}=110\text{atm}+45.6\text{atm}$

Therefore, $P_{\text{total}}=155.6\text{atm}$.

Solve for the partial pressure of each reactant at equilibrium by substituting the calculated $P_{\text{total}}$ and x.

$P_{{\text{H}}_{2,\text{m}}}=0.75\cdot P_{\text{total}}-3x$

$\left(0.75\right)(155.6\text{atm})-3(22.8\text{atm})$

$P_{{\text{H}}_{2,\text{m}}}=48.3\text{atm}$

$P_{{\text{N}}_{2,\dots 9}}=0.25\cdot P_{\text{total}}-x$

$=\left(0.25\right)(155.6\text{atm})-22.8\text{atm}$

.$P_{{\text{N}}_{2,\text{m}}}=16.1\text{atm}$

Next, write the expression for the equilibrium constant of the reaction in terms of partial pressures.

$K_p=\frac{P_{\text{products}}}{P_{\text{reactants}}}$

$K_p=\frac{P_{{\text{NH}}_3}^2}{P_{{\text{H}}_2}^3\cdot P_{{\text{N}}_2}}$

Lastly, solve for$K_p$ by substituting the values.

$K_p=\frac{P_{{\text{NH}}_3}^2}{P_{{\text{H}}_2}^3\cdot P_{{\text{N}}_2}}$

$=\frac{{\left(45.6\right)}^2}{{\left(48.3\right)}^3\left(16.1\right)}$

$K_p=1.15\times {10}^{-3}$

Therefore, the required value is $K_p=1.15\times {10}^{-3}$.