Question

1. at 500 k, a 10.0 l equilibrium mixture contains 0.424 mol n₂, 1.272 mol h₂, and 1.152 mol nh₃. the mixture is quickly chilled to a temperature at which the nh₃ liquefies, and the nh₃(l) is completely removed. the 10.0 l gaseous mixture is then returned to 500 k, and equilibrium is re - established. how many moles of nh₃ will be present in the new equilibrium mixture?

Explanation:

Step1: Write the reaction equation

$N_2(g)+3H_2(g)
ightleftharpoons 2NH_3(g)$

Step2: Calculate the initial equilibrium concentrations

$[N_2]_{eq1}=\frac{0.424\ mol}{10.0\ L}=0.0424\ M$, $[H_2]_{eq1}=\frac{1.272\ mol}{10.0\ L}=0.1272\ M$, $[NH_3]_{eq1}=\frac{1.152\ mol}{10.0\ L}=0.1152\ M$

Step3: Calculate the equilibrium constant $K_c$

$K_c=\frac{[NH_3]_{eq1}^2}{[N_2]_{eq1}[H_2]_{eq1}^3}=\frac{(0.1152)^2}{0.0424\times(0.1272)^3}$

Step4: After removing $NH_3$, find new initial concentrations

$[N_2]_{init2}=0.0424\ M$, $[H_2]_{init2}=0.1272\ M$, $[NH_3]_{init2}=0\ M$
Let $x$ be the change in concentration of $N_2$ at new - equilibrium. Then the equilibrium concentrations are $[N_2]_{eq2}=0.0424 - x$, $[H_2]_{eq2}=0.1272-3x$, $[NH_3]_{eq2}=2x$

Step5: Substitute into the $K_c$ expression

$K_c=\frac{(2x)^2}{(0.0424 - x)(0.1272 - 3x)^3}$
Since $K_c$ is a constant and we know its value from Step 3, we can solve for $x$ (using an iterative or approximation method if needed). For a small - change approximation (if valid, which we can check later), assume $0.0424 - x\approx0.0424$ and $0.1272-3x\approx0.1272$
$K_c=\frac{(2x)^2}{0.0424\times(0.1272)^3}$
Solve for $x$ and then find $[NH_3]_{eq2}=2x$. Multiply by the volume ($V = 10.0\ L$) to get the number of moles of $NH_3$.
$K_c=\frac{(0.1152)^2}{0.0424\times(0.1272)^3}\approx17.0$
$\frac{(2x)^2}{0.0424\times(0.1272)^3}=17.0$
$(2x)^2=17.0\times0.0424\times(0.1272)^3$
$(2x)^2=17.0\times0.0424\times0.00205$
$(2x)^2 = 0.00144$
$2x=\sqrt{0.00144}\approx0.038$
The number of moles of $NH_3$ in the new equilibrium mixture $n = 2x\times V$
$n=0.038\ M\times10.0\ L = 0.38\ mol$

Answer:

$0.38\ mol$