1.0 mol of CO and 2.0 mol of H2 are heated in a 2.0 L rigid vessel at 500 K. At equilibrium of CO(g) + 2H2(g) CH3OH(g), 0.30 mol of CH3OH is present. Using R = 0.0821 L atm mol^-1 K^-1, calculate K_p in units of 10^-4 atm^-2, correct to two decimal places (give the number that multiplies 10^-4).

JEE Advanced levelNumerical answerDifficulty 3 of 3ce097

1.0 mol of $CO$ and 2.0 mol of $H X_{2}$ are heated in a 2.0 L rigid vessel at 500 K. At equilibrium of $CO (g) + 2 H X_{2} (g) CH X_{3} OH (g)$ , 0.30 mol of $CH X_{3} OH$ is present. Using $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , calculate $K_{p}$ in units of $1 0^{- 4} at m^{- 2}$ , correct to two decimal places (give the number that multiplies $1 0^{- 4}$ ).

Numerical answer type, answer in 10^-4 atm^-2, tolerance 0.01.

Equilibrium moles:

CO

0.70,

H X_{2}

1.40,

CH X_{3} OH

0.30. Concentrations (divide by 2.0 L): 0.35, 0.70, 0.15 mol/L.

K_{c} = 0.15/ (0.35 \times 0.7 0^{2}) = 0.8746 L^{2} mo l^{- 2}

. With

Δ n = - 2

K_{p} = K_{c} / (RT)^{2} = 0.8746/ (41.05)^{2} = 5.190 \times 1 0^{- 4} at m^{- 2}