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प्रश्न
For the reaction, \[\mathrm{N_{2(g)}+3H_{2(g)}\longrightarrow2NH_{3(g)}}\]
N2(g) disappears at a rate of 2×10-4mol dm-3 s-1. Calculate rate of disappearance of H2?
पर्याय
\[1\times10^{-4}\mathrm{~mol~dm^{-3}~s^{-1}}\]
\[6.0\times10^{-4}moldm^{-3}s^{-1}\]
\[2.0\times10^{-4}\mathrm{~mol~dm^{-3}~s^{-1}}\]
\[2.5\times10^{-4}\mathrm{~mol~dm}^{-3}\mathrm{~s}^{-1}\]
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उत्तर
\[6.0\times10^{-4}moldm^{-3}s^{-1}\]
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
\[=\frac{-\Delta\left[\mathrm{H}_2\right]}{\Delta\mathrm{t}}=3\times\left(\frac{-\Delta\left[\mathrm{N}_2\right]}{\Delta\mathrm{t}}\right)\]
\[=3\times2\times10^{-4}=6.0\times10^{-4}\mathrm{~mol~dm}^{-3}\mathrm{s}^{-1}\]
