Question

a chemical engineer is studying the following reaction: n₂(g)+3h₂(g)→2nh₃(g). at the temperature the engineer picks, the equilibrium constant kₚ for this reaction is 0.0049. the engineer changes (\fills\) three reaction vessels with nitrogen and hydrogen, and lets the reaction begin. he then measures the composition of the mixture inside each vessel from time to time. his first set of measurements are shown in the table below. predict the changes in the compositions the engineer should expect next time he measures the compositions.

compound	n₂	h₂	nh₃
expected change in pressure	1 increase	1 increase	1 decrease	1 decrease	1 decrease	1 increase	1 increase	1 increase	(no change)	(no change)	(no change)	(no change)

Explanation:

Step1: Write the expression for the reaction - quotient $Q_p$

The reaction is $N_2(g)+3H_2(g)
ightleftharpoons 2NH_3(g)$, and the expression for the reaction - quotient $Q_p=\frac{P_{NH_3}^2}{P_{N_2}\times P_{H_2}^3}$, where $P_{NH_3}$, $P_{N_2}$, and $P_{H_2}$ are the partial pressures of $NH_3$, $N_2$, and $H_2$ respectively. Given $P_{N_2} = 37.22\ atm$, $P_{H_2}=23.66\ atm$, and $P_{NH_3}=49.13\ atm$.

Step2: Calculate $Q_p$

$Q_p=\frac{(49.13)^2}{37.22\times(23.66)^3}=\frac{2413.7569}{37.22\times13392.97}=\frac{2413.7569}{498496.3434}\approx0.00484$.

Step3: Compare $Q_p$ with $K_p$

The equilibrium constant $K_p = 0.0049$. Since $Q_p(0.00484)

Step4: Determine the changes in partial - pressures

When the reaction shifts to the right, the amount of $N_2$ and $H_2$ will decrease (because they are reactants being consumed), and the amount of $NH_3$ will increase (because it is a product being formed). Also, according to Le - Chatelier's principle, as the reaction proceeds to the right, the total number of moles of gas decreases (4 moles of reactants produce 2 moles of products), so the pressure will decrease.

Answer:

$NH_3$: increase
$H_2$: decrease
$N_2$: decrease
pressure: decrease