The half-life for the first-order decomposition of nitramide: NH2NO2(aq) → N2O(g) + H2O(l) is 123 min at 27°C. If 200 ml of 0.02 M NH2NO2 solution is allowed to decompose, how long (in min) must the r

Question

The half-life for the first-order decomposition of nitramide: NH2NO2(aq) → N2O(g) + H2O(l) is 123 min at 27°C. If 200 ml of 0.02 M NH2NO2 solution is allowed to decompose, how long (in min) must the reaction proceed to produce 49.26 ml of 'wet' N2O(g) measured at 27°C and 785 mm Hg? The vapour pressure of water at 27°C is 25 mm Hg.

Prashant Kotian · Accepted Answer

Correct answer: 0123. P_N2O = 785 - 25 = 760 mm = 1 atm. Moles of N2O = PV/RT = (1 × 0.04926) / (0.0821 × 300) = 0.002 mol. Initial moles of nitramide = 200 ml × 0.02 M = 0.004 mol. 0.002 moles of N2O means 0.002 moles of nitramide reacted, which is exactly 50% of the initial amount. The time required is exactly one half-life, which is 123 min. Therefore, correct answer is 0123.

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