At $$ 25^{\circ} \mathrm{C} $$, the reaction $$ \begin{array}{l} 2 \mathrm{NH}_{3}(g) \rightleftarrows \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \ \text { has } K_{\mathrm{c}}=2.3 \times 10^{-9} \end{array} $$ If $$ 0.062 \mathrm{~mol} \mathrm{NH}_{3} $$ is placed in a 3.61 L container, what will the concentrations of $$ \mathrm{N}_{2} $$ and $$ \mathrm{H}_{2} $$ be when equilibrium is established? Make simplifying assumptions in your calculations. Assume the change in $$ \mathrm{NH}_{3} $$ concentration is insignificant if compared to initial value. Incorrect. $$ \left[N_{2}\right]=\mathbf{i} \quad!M $$ eTextbook and Media Incorrect. $$ \left[H_{2}\right]= $$ $$ \square $$ M

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We start with the reaction

$$
2\,\mathrm{NH}_3(g) \rightleftharpoons \mathrm{N}_2(g) + 3\,\mathrm{H}_2(g)
$$

with

$$
K_c = 2.3\times10^{-9}.
$$

A total of $$0.062~\mathrm{mol}$$ of $$\mathrm{NH}_3$$ is placed in a $$3.61~\mathrm{L}$$ container. Therefore, the initial concentration of $$\mathrm{NH}_3$$ is

$$
[\mathrm{NH}_3]_{\text{initial}} = \frac{0.062~\mathrm{mol}}{3.61~\mathrm{L}} \approx 0.01716~\mathrm{M}.
$$

It is stated that we can assume the change in the $$\mathrm{NH}_3$$ concentration is insignificant. Let the reaction produce a small amount $$x$$ (in moles) of $$\mathrm{N}_2$$ at equilibrium. The stoichiometric relationships are:

- For every $$2$$ moles of $$\mathrm{NH}_3$$ that react, $$1$$ mole of $$\mathrm{N}_2$$ is produced.
- For every $$2$$ moles of $$\mathrm{NH}_3$$ that react, $$3$$ moles of $$\mathrm{H}_2$$ are produced.

Thus, if $$x$$ moles of $$\mathrm{N}_2$$ are produced then:

$$
\begin{array}{rcl}
\text{Change in } \mathrm{NH}_3 & = & -2x, \
\text{Change in } \mathrm{N}_2 & = & +x, \
\text{Change in } \mathrm{H}_2 & = & +3x.
\end{array}
$$

At equilibrium the concentrations are:

$$
\begin{aligned}
[\mathrm{NH}_3] &\approx 0.01716~\mathrm{M} \quad (\text{since } 2x \text{ is negligible}),\[1mm]
[\mathrm{N}_2] &= \frac{x}{3.61}, \[1mm]
[\mathrm{H}_2] &= \frac{3x}{3.61}.
\end{aligned}
$$

The equilibrium constant expression for the reaction is

$$
K_c = \frac{[\mathrm{N}_2][\mathrm{H}_2]^3}{[\mathrm{NH}_3]^2}.
$$

Substitute the equilibrium concentrations:

$$
K_c = \frac{\left(\dfrac{x}{3.61}\right)\left(\dfrac{3x}{3.61}\right)^3}{(0.01716)^2}.
$$

First, compute $$\left(\frac{3x}{3.61}\right)^3$$:

$$
\left(\frac{3x}{3.61}\right)^3 = \frac{27x^3}{3.61^3}.
$$

Thus, the expression becomes:

$$
K_c = \frac{\dfrac{x}{3.61}\cdot \dfrac{27x^3}{3.61^3}}{(0.01716)^2} = \frac{27\,x^4}{3.61^4\,(0.01716)^2}.
$$

Now, substitute $$K_c = 2.3\times10^{-9}$$:

$$
\frac{27\,x^4}{3.61^4\,(0.01716)^2} = 2.3\times10^{-9}.
$$

Solve for $$x^4$$:

$$
x^4 = \frac{2.3\times10^{-9}\times3.61^4\times (0.01716)^2}{27}.
$$

Let’s compute the numerical values step by step.

1. Compute $$(0.01716)^2$$:

$$
(0.01716)^2 \approx 0.0002946.
$$

2. Compute $$3.61^4$$:

$$
3.61^2 \approx 13.0321,\quad 3.61^4 \approx (13.0321)^2 \approx 169.83.
$$

3. Multiply $$3.61^4$$ and $$(0.01716)^2$$:

$$
169.83\times 0.0002946 \approx 0.0500.
$$

4. Multiply by $$2.3\times10^{-9}$$:

$$
2.3\times10^{-9} \times 0.0500 \approx 1.15\times10^{-10}.
$$

5. Divide by 27:

$$
x^4 \approx \frac{1.15\times10^{-10}}{27} \approx 4.26\times10^{-12}.
$$

Now, take the fourth root:

$$
x = \sqrt[4]{4.26\times10^{-12}}.
$$

Taking the fourth root means taking the square root twice. First, take the square root:

$$
\sqrt{4.26\times10^{-12}} \approx 2.064\times10^{-6}, 
$$
since $$(2.064\times10^{-6})^2 \approx 4.26\times10^{-12}$$.

Then, take the square root again:

$$
\sqrt{2.064\times10^{-6}} \approx 1.437\times10^{-3}.
$$

Thus, we find

$$
x \approx 1.44\times10^{-3}~\mathrm{mol}.
$$

Recall that

$$
[\mathrm{N}_2] = \frac{x}{3.61} \quad \text{and} \quad [\mathrm{H}_2] = \frac{3x}{3.61}.
$$

Compute these:

$$
[\mathrm{N}_2] \approx \frac{1.44\times10^{-3}}{3.61} \approx 3.99\times10^{-4}~\mathrm{M},
$$

$$
[\mathrm{H}_2] \approx \frac{3(1.44\times10^{-3})}{3.61} \approx 1.20\times10^{-3}~\mathrm{M}.
$$

Thus, the equilibrium concentrations are:

$$
\boxed{[\mathrm{N}_2] \approx 4.0\times10^{-4}~\mathrm{M},\quad [\mathrm{H}_2] \approx 1.2\times10^{-3}~\mathrm{M}.}
$$
At equilibrium, the concentrations are: $$ \left[\mathrm{N}_2\right] \approx 4.0 \times 10^{-4}~\mathrm{M}, \quad \left[\mathrm{H}_2\right] \approx 1.2 \times 10^{-3}~\mathrm{M}. $$

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