Question

For the reaction N2() + 3H2(g) → 2NH3(g) under certain conditions of temperature and partial pressure of the reactants, the rate
of formation of NH , is 0.001 kg h". The rate of conversion of H, under the same conditions is first change i
(a) 1.82 * 10-4 kg/hr (b) 0.0015 kg Ihr (@) 1.52 * 104 kg / hr (d) 1.82 x 10-14 kg / hr​

Answer 1

Answer : The correct option is, (b) 0.0015 kg/hr

Explanation : Given,

$\frac{d[NH_3]}{dt}$ = 0.001 kg/hr

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

The rate of disappearance of $N_2$ = $-\frac{d[N_2]}{dt}$

The rate of disappearance of $H_2$ = $-\frac{1}{3}\frac{d[H_2]}{dt}$

The rate of formation of $NH_3$ = $+\frac{1}{2}\frac{d[NH_3]}{dt}$

As we know that,

$\frac{d[NH_3]}{dt}$ = 0.001kg/hr

So,

$-\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{1}{2}\frac{d[NH_3]}$

$-\frac{d[H_2]}{dt}=\frac{3}{2}\times \frac{1}{2}\frac{d[NH_3]}$

$-\frac{d[H_2]}{dt}=\frac{3}{2}\times 0.001kg/hr=0.0015kg/hr$  

Therefore, the rate of conversion of H, under the same conditions is 0.0015 kg/hr

Answer 2

Answer:

Explanation:the answer is. 0.176*10^-3