The decomposition of a gas on a catalyst surface is zero order, and the pressure of the gas falls at a constant 8.0x10^-3 atm min^-1 at 400 K. Treating the gas as ideal (R = 0.0821 L atm mol^-1 K^-1), find the zero order rate constant in concentration terms, in units of 10^-6 mol L^-1 s^-1 (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ck053

The decomposition of a gas on a catalyst surface is zero order, and the pressure of the gas falls at a constant $8.0 \times 1 0^{- 3} atm mi n^{- 1}$ at $400 K$ . Treating the gas as ideal ( $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ ), find the zero order rate constant in concentration terms, in units of $1 0^{- 6} mol L^{- 1} s^{- 1}$ (two decimals).

Numerical answer type, answer in 10^-6 mol L^-1 s^-1, tolerance 0.02.

Since

c = p / RT

, the rate in concentration terms is

k_{p} / RT

with

RT = 0.0821 \times 400 = 32.84 L atm mo l^{- 1}

8.0 \times 1 0^{- 3} /32.84 = 2.4361 e - 04 mol L^{- 1} mi n^{- 1}

. Dividing by 60 to convert to seconds gives

4.06 \times 1 0^{- 6} mol L^{- 1} s^{- 1}