Q17.103P - Chemistry: Molecular Nature Of Matter And Change

Question

Mixture of  $3.00$ volumes of ${\text{H}}_2$ and $1.00$ volume of  ${\text{N}}_2$ reacts at  ${344}^{°}\text{C}$  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.

Step-by-Step Solution

Answer

 for the reaction is  $K_p=1.15\times {10}^{-3}$.

Step 1: Definition of volume

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

Step 2: Solve the total volume of the system

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.

 $V_{\text{total }}={\text{ volume of H}}_2+{\text{ volume of N}}_2$

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

Therefore the total volume is .$4.00$

Step 3: Write the reaction table using the given values

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.

Step 4: Equate the equilibrium formula

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

Step 5: Solve for K p

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  $K_p=1.15\times {10}^{-3}$.