Chemical Kinetics - Result Question 10
####10. For the reaction, $N_2+3 H_2 \longrightarrow 2 NH_3$,
[2009]
$\frac{d[NH_3]}{d t}=2 \times 10^{-4} mol L^{-1} s^{-1}$, the value of
$\frac{-d[H_2]}{d t}$ would be:
(a) $4 \times 10^{-4} mol L^{-1} s^{-1}$
(b) $6 \times 10^{-4} mol L^{-1} s^{-1}$
(c) $1 \times 10^{-4} mol L^{-1} s^{-1}$
(d) $3 \times 10^{-4} mol L^{-1} s^{-1}$
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Solution:
- (d) Rate of disappearance of $H_2=$ rate of formation of $NH_3$.
$-\frac{1}{3} \frac{d[H_2]}{d t}=\frac{1}{2} \frac{d[NH_3]}{d t}$
$\Rightarrow \frac{-d[H_2]}{d t}=\frac{3}{2} \frac{d[NH_3]}{d t}=\frac{3}{2} \times 2 \times 10^{-4}$
$=3 \times 10^{-4} mol L^{-1} s^{-1}$
