Question

For the reaction $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$, under certain conditions of temperature and partial pressure of the reactants, the rate of formation of $N{H}_3$ is $1.0\times {10}^{-4}molli{t}^{-1}{h}^{-1}$. The rate of conversion of $H_2$ under the same conditions is:

• A. $6.7\times {10}^{-4}molli{t}^{-1}{h}^{-1}$
• B. $3.3\times {10}^{-3}molli{t}^{-1}{h}^{-1}$
• C. $1.5\times {10}^{-4}molli{t}^{-1}{h}^{-1}$
• D. $0.5\times {10}^{-4}molli{t}^{-1}{h}^{-1}$

Answer 1

The correct option is C $1.5\times {10}^{-4}molli{t}^{-1}{h}^{-1}$

$N_2\left(g\right)+3H_2\left(g\right)⟶2N{H}_3\left(g\right)$

Rate of $R_{\alpha }^n\left(ROR\right)=\frac{-1}{1}\frac{d\left[N_{2\left(g\right)}\right]}{dt}=\frac{-1}{3}\frac{d\left[H_{2\left(g\right)}\right]}{dt}=\frac{1}{2}\frac{d\left[N{H}_{3\left(g\right)}\right]}{dt}$

$ROR=\frac{RODof{N}_{2\left(g\right)}}{1}=\frac{RODof{H}_{2\left(g\right)}}{3}=\frac{ROAofN{H}_{3\left(g\right)}}{2}$

$ROA=$ of $N{H}_{3\left(g\right)}=2\times ROR\Rightarrow ROR=\frac{1\times {10}^{-4}}{2}$

$ROD=$ Rate of disappearance

$ROA=$ Rate of Appearance.

$ROR=\frac{1\times {10}^{-4}}{2}=\frac{RODof{H}_2}{3}$

$RODof{H}_2=\frac{3}{2}\times {10}^{-4}molli{t}^{-1}{h}^{-1}$