MHT CET · Chemistry · Chemical Kinetics
For the reaction,
\(
2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})} \text { If } \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}\) \(=0.052 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}
\)
Calculate rate of consumption of \(\mathrm{NO}_{(\mathrm{g})}\).
- A \(0.114 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
- B \(0.078 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
- C \(0.026 \mathrm{~mol}^{-3} \mathrm{~s}^{-1}\)
- D \(0.052 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(D) \(0.052 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
From reaction rate expression is
\(\begin{aligned} & -\frac{1}{2} \frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{dt}}=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \\ & \therefore-\frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}\end{aligned}\)
\(\begin{aligned} & -\frac{1}{2} \frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{dt}}=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \\ & \therefore-\frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}\end{aligned}\)
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