Ammonia decomposes on a hot tungsten surface by a zero order reaction, 2NH3(g) -> N2(g) + 3H2(g), at constant volume and temperature. Starting with pure NH3 at 200 mmHg, the total pressure is 233 mmHg after 30 min. Calculate the time (in minutes, two decimals) for the partial pressure of NH3 to fall to 25% of its initial value.

JEE Advanced levelNumerical answerDifficulty 3 of 3ck107

Ammonia decomposes on a hot tungsten surface by a zero order reaction, $2 NH X_{3} (g) N X_{2} (g) + 3 H X_{2} (g)$ , at constant volume and temperature. Starting with pure $NH X_{3}$ at $200 mmHg$ , the total pressure is $233 mmHg$ after $30 min$ . Calculate the time (in minutes, two decimals) for the partial pressure of $NH X_{3}$ to fall to 25% of its initial value.

Numerical answer type, answer in min, tolerance 0.1.

Every

NH X_{3}

consumed raises the total pressure by 1 unit (2 mol gives 4 mol), so

p (NH X_{3}) = 2 P_{0} - P_{t} = 400 - 233 = 167 mmHg

at 30 min. The zero order rate is

k = (200 - 167) /30 = 1.1 mmHg mi n^{- 1}

. A fall to 25% means a loss of 150 mmHg, so

t = 150/1.1 = 136.36 min