3. nh₃ is normally encountered as a gas with a pungent odor. it is formed from hydrogen and nitrogen…

3. nh₃ is normally encountered as a gas with a pungent odor. it is formed from hydrogen and nitrogen (equation given below). a chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0 litre reaction vessel in 1:3 ratio. the equilibrium constant for the reaction is 1.0 × 10⁻⁴, and the equilibrium value of h₂(g) is 0.12 mol/l. calculate the equilibrium value of nh₃.\n\nn₂(g) + 3 h₂(g) ⇔ 2 nh₃(g)

3. nh₃ is normally encountered as a gas with a pungent odor. it is formed from hydrogen and nitrogen (equation given below). a chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0 litre reaction vessel in 1:3 ratio. the equilibrium constant for the reaction is 1.0 × 10⁻⁴, and the equilibrium value of h₂(g) is 0.12 mol/l. calculate the equilibrium value of nh₃.\n\nn₂(g) + 3 h₂(g) ⇔ 2 nh₃(g)

Answer

Explanation:

Step1: Recall Equilibrium Constant Formula

For the reaction $\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)}$, the equilibrium constant expression $K_c$ is $K_c = \frac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}$. First, find $[\ce{N2}]$ using the initial ratio and equilibrium $[\ce{H2}]$. The initial ratio of $\ce{N2}:\ce{H2}$ is $1:3$, so at equilibrium, the ratio should still hold (since the stoichiometry is $1:3$ for reactants). Given $[\ce{H2}] = 0.12\ \text{mol/L}$, then $[\ce{N2}] = \frac{[\ce{H2}]}{3} = \frac{0.12}{3} = 0.04\ \text{mol/L}$.

Step2: Substitute Values into $K_c$ Formula

We know $K_c = 1.0 \times 10^{-4}$, $[\ce{H2}] = 0.12\ \text{mol/L}$, $[\ce{N2}] = 0.04\ \text{mol/L}$. Substitute into $K_c$:
$1.0 \times 10^{-4} = \frac{[\ce{NH3}]^2}{(0.04)(0.12)^3}$

Step3: Solve for $[\ce{NH3}]^2$

First calculate the denominator: $(0.04)(0.12)^3 = 0.04 \times 0.001728 = 6.912 \times 10^{-5}$.
Then, $[\ce{NH3}]^2 = K_c \times (0.04)(0.12)^3 = 1.0 \times 10^{-4} \times 6.912 \times 10^{-5} = 6.912 \times 10^{-9}$.

Step4: Solve for $[\ce{NH3}]$

Take the square root of $6.912 \times 10^{-9}$: $[\ce{NH3}] = \sqrt{6.912 \times 10^{-9}} \approx \sqrt{69.12 \times 10^{-10}} = \sqrt{69.12} \times 10^{-5} \approx 8.314 \times 10^{-5}\ \text{mol/L}$ (or more precisely, $\sqrt{6.912 \times 10^{-9}} \approx 8.31 \times 10^{-5}\ \text{mol/L}$).

Answer:

The equilibrium value of $[\ce{NH3}]$ is approximately $\boldsymbol{8.3 \times 10^{-5}\ \text{mol/L}}$ (or more precisely $8.31 \times 10^{-5}\ \text{mol/L}$).