The reaction \(\mathrm{2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)}\) follows first order kinetics. The pressure of a vessel containing only \(\mathrm{N_2O_5}\) was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. Will be (Assume temperature remains constant) :
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- 106.25 mm Hg
- 125 mm Hg
- 116.25 mm Hg
- 150 mm Hg
The Correct Option is A
Solution and Explanation
At 30 min: \(50 + 3p = 87.5\), so \(p = 12.5\); \(\mathrm{N_2O_5}\) left = \(50 - 25 = 25\) mm Hg, half of the initial, so \(t_{1/2} = 30\) min.
At 60 min (two half-lives): \(\mathrm{N_2O_5}\) left = 12.5 mm Hg, consumed = 37.5 so \(2p = 37.5\), \(p = 18.75\).
\(P_{total} = 50 + 3(18.75) = 106.25\) mm Hg.
Answer: (1) 106.25 mm Hg
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