Chemical Kinetics and Nuclear ChemistryHard

Question

The decomposition of N2O5 according to the equation 2N2O5(g) → 4NO2(g) + O2(g) is a first order reaction. After 30 minutes, from the start of the decomposition in a closed vessel, the total pressure developed is found to be 300 mm of Hg and on complete decomposition, the total pressure is 600 mm of Hg. The rate constant of the reaction is (ln 1.2 = 0.18)

Options

A.6.0 × 10−3 min−1
B.0.18 min−1
C.6.0 × 10−2 min−1
D.3.0 × 10−3 min−1

Solution

2N2O5 ? 4NO2 + O2

t = 0 P0 0 0

t = 30 min P0 – x 2x x/2

t = ∞ ≈ 0 2P0 P0/2

From question: $2P_{0} + \frac{P_{0}}{2} = 600$

$\Rightarrow P_{0} = 240\text{ mm}$

and $\left( P_{0} - x \right) + 2x + \frac{x}{2} = 300 \Rightarrow x = 40\text{ min}$

Now, $K_{N_{2}O_{5}} = \frac{1}{t}.\ln\frac{P_{N_{2}O_{5}}^{o}}{P_{N_{2}O_{5}}} = \frac{1}{30\text{ min}}.\ln\frac{240}{240 - 40} = 6 \times 10^{- 3}\text{ mi}\text{n}^{- 1}$

$\therefore K_{rxn} = \frac{K_{N_{2}O_{5}}}{2} = 3 \times 10^{- 3}\text{mi}\text{n}^{- 1}$

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