In the synthesis of $\text{NH}_3$ from $\text{H}_2$ and $\text{N}_2$ if $6\times10^{-2}$ mole of hydrogen disappears in 10 minutes, the number of moles of $\text{NH}_3$ formed in 3 minutes is:
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To simplify tracking multi-step kinetics, convert everything to a $1$-minute baseline first! $6\times10^{-2}$ over $10$ minutes is $6\times10^{-3}$ per minute. Multiplying by $0.3$ minutes gives $1.8\times10^{-3}$ moles of $\text{H}_2$. Applying the $\frac{2}{3}$ multiplier directly yields $1.2\times10^{-3}$ moles of $\text{NH}_3$ instantly.
Understanding the Concept:
The balanced chemical equation for Haber's process is:
\[
\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)
\]
According to the stoichiometry of the reaction, the rate of disappearance of hydrogen and the rate of appearance of ammonia are related by:
\[
-\frac{1}{3}\frac{\Delta n_{\text{H}_2}}{\Delta t} = +\frac{1}{2}\frac{\Delta n_{\text{NH}_3}}{\Delta t} \implies \Delta n_{\text{NH}_3} = \frac{2}{3} \left(-\Delta n_{\text{H}_2}\right)
\]
Step 1: Calculate the rate of disappearance of hydrogen per minute.
We are given that $\Delta n_{\text{H}_2} = 6\times10^{-2}$ moles disappear in $\Delta t = 10$ minutes.
\[
\text{Rate of disappearance of }\text{H}_2 = \frac{6\times10^{-2}\text{ moles}}{10\text{ min}} = 6\times10^{-3}\text{ mol min}^{-1}
\]
Step 2: Determine the number of moles of hydrogen that disappear in 0.3 minutes.
For a time interval of $t = 0.3$ minutes:
\[
\text{Moles of }\text{H}_2\text{ disappeared} = (6\times10^{-3}\text{ mol min}^{-1}) \times 0.3\text{ min} = 1.8\times10^{-3}\text{ moles}
\]
Step 3: Use stoichiometry to find the moles of $\text{NH}_3$ produced.
From the balanced equation, 3 moles of $\text{H}_2$ produce 2 moles of $\text{NH}_3$. Therefore:
\[
\text{Moles of }\text{NH}_3\text{ formed} = \frac{2}{3} \times (\text{moles of }\text{H}_2\text{ disappeared})
\]
\[
\text{Moles of }\text{NH}_3\text{ formed} = \frac{2}{3} \times 1.8\times10^{-3} = 2 \times 0.6\times10^{-3} = 1.2\times10^{-3}\text{ moles}
\]