Question

For the reaction, \[\mathrm{N_{2(g)}+3H_{2(g)}\longrightarrow2NH_{3(g)}}\]
N_2(g) disappears at a rate of 2×10^-4mol dm^-3 s^-1. Calculate rate of disappearance of H₂?

Options

• \[1\times10^{-4}\mathrm{~mol~dm^{-3}~s^{-1}}\]

• \[6.0\times10^{-4}moldm^{-3}s^{-1}\]

• \[2.0\times10^{-4}\mathrm{~mol~dm^{-3}~s^{-1}}\]

• \[2.5\times10^{-4}\mathrm{~mol~dm}^{-3}\mathrm{~s}^{-1}\]

Answer 1

\[6.0\times10^{-4}moldm^{-3}s^{-1}\]

Explanation:

\[\mathrm{N}_2+3\mathrm{H}_2\longrightarrow2\mathrm{NH}_3\]
\[\mathrm{Rate}=\frac{-\Delta\left[\mathrm{N}_2\right]}{\Delta\mathrm{t}}=\frac{1}{3}\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta\mathrm{t}}=\frac{1}{2}\frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta\mathrm{t}}\]
∴ Rate of disappearance of H₂
\[=\frac{-\Delta\left[\mathrm{H}_2\right]}{\Delta\mathrm{t}}=3\times\left(\frac{-\Delta\left[\mathrm{N}_2\right]}{\Delta\mathrm{t}}\right)\] 
\[=3\times2\times10^{-4}=6.0\times10^{-4}\mathrm{~mol~dm}^{-3}\mathrm{s}^{-1}\]