In a catalytic experiment involving the Haber process, , the rate of appearance of is measured as . — Chemical Kinetics Chemistry Question
Question
In a catalytic experiment involving the Haber process, $N_2 + 3H_2 \rightarrow 2NH_3$, the rate of appearance of $NH_3$ is measured as $2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$. What is the rate of disappearance of $H_2$ in $\text{mol L}^{-1} \text{ s}^{-1}$? Express your answer as $X \times 10^{-4}$ and provide the value of $X$.
Answer: 3.75
💡 Solution & Explanation
Rate $= -\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt}$. $-\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt} = \frac{3}{2} \times 2.5 \times 10^{-4} = 3.75 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$. Thus, $X = 3.75$.
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