Consider the following equilibrium, mathrmCO(mathrmg) + 2mathrmH_2(mathrmg) rightleftharpoons mathrmCH_3mathrmOH(mathrmg)$$\mathrm{CO}(\mathrm{g}) + 2\mathrm{H}_2(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3\mathrm{OH}(\mathrm{g})$$ 0.1 mol of CO along with a catalyst is present in a 2mathrmdm^3$2\mathrm{dm}^3$ flask maintained at 500mathrmK$500\mathrm{K}$. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of mathrmCH_3mathrmOH$\mathrm{CH}_3\mathrm{OH}$ is formed. The mathrmK_p^0$\mathrm{K_p^0}$ is times 10^-3$\times 10^{-3}$ (nearest integer). Given: mathrmR = 0.08mathrmdm^3cdottextbarcdotmathrmK^-1cdottextmol^-1$\mathrm{R} = 0.08\mathrm{dm}^3\cdot\text{bar}\cdot\mathrm{K}^{-1}\cdot\text{mol}^{-1}$ Assume only methanol is formed as the product and the system follows ideal gas behaviour.

### Related Formula Ideal Gas equation layout for aggregate systems: P_texttotal cdot V = n_texttotal cdot RT$$P_{\text{total}} \cdot V = n_{\text{total}} \cdot RT$$ Partial Pressure expression using mole fractions: p_i = X_i cdot P_texttotal$$p_i = X_i \cdot P_{\text{total}}$$ ### Core Logic Let's tabulate equilibrium progress row-by-row: * Reaction matrix: beginarraylcccc & mathrmCO(g) & + & mathrm2H_2(g) & rightleftharpoons & mathrmCH_3OH(g) \\ t=0 & 0.1 & & a & & 0 \\ t_texteq & 0.1 - x & & a - 2x & & x endarray$$\begin{array}{lcccc} & \mathrm{CO(g)} & + & \mathrm{2H_2(g)} & \rightleftharpoons & \mathrm{CH_3OH(g)} \\ t=0 & 0.1 & & a & & 0 \\ t_{\text{eq}} & 0.1 - x & & a - 2x & & x \end{array}$$ * Given x = 0.04mathrm~mol$x = 0.04\mathrm{~mol}$ at equilibrium: - n_mathrmCO = 0.1 - 0.04 = 0.06mathrm~mol$n_{\mathrm{CO}} = 0.1 - 0.04 = 0.06\mathrm{~mol}$ - n_mathrmCH_3OH = 0.04mathrm~mol$n_{\mathrm{CH_3OH}} = 0.04\mathrm{~mol}$ * Determine total moles via system pressure (P = 5mathrm~bar$P = 5\mathrm{~bar}$, V = 2mathrm~L$V = 2\mathrm{~L}$, T = 500mathrmK$T = 500\mathrm{K}$): 5 times 2 = n_texttotal times 0.08 times 500 implies n_texttotal = frac1040 = 0.25mathrm~mol$$5 \times 2 = n_{\text{total}} \times 0.08 \times 500 \implies n_{\text{total}} = \frac{10}{40} = 0.25\mathrm{~mol}$$ * Find remaining unknown hydrogen moles: n_texttotal = 0.06 + n_mathrmH_2 + 0.04 = 0.25 implies n_mathrmH_2 = 0.15mathrm~mol$$n_{\text{total}} = 0.06 + n_{\mathrm{H_2}} + 0.04 = 0.25 \implies n_{\mathrm{H_2}} = 0.15\mathrm{~mol}$$ ### Step 1: Calculate Kp Compute partial pressures using fractional allocation fractions (n_texttotal = 0.25$n_{\text{total}} = 0.25$): * p_mathrmCH_3OH = frac0.040.25 times 5 = 0.8mathrm~bar$p_{\mathrm{CH_3OH}} = \frac{0.04}{0.25} \times 5 = 0.8\mathrm{~bar}$ * p_mathrmCO = frac0.060.25 times 5 = 1.2mathrm~bar$p_{\mathrm{CO}} = \frac{0.06}{0.25} \times 5 = 1.2\mathrm{~bar}$ * p_mathrmH_2 = frac0.150.25 times 5 = 3.0mathrm~bar$p_{\mathrm{H_2}} = \frac{0.15}{0.25} \times 5 = 3.0\mathrm{~bar}$ Substitute these pressures into the equilibrium expression: K_p = fracp_mathrmCH_3OHp_mathrmCO cdot (p_mathrmH_2)^2 = frac0.81.2 times 3^2 = frac0.810.8 = 0.07407 = 74.07 times 10^-3$$K_p = \frac{p_{\mathrm{CH_3OH}}}{p_{\mathrm{CO}} \cdot (p_{\mathrm{H_2}})^2} = \frac{0.8}{1.2 \times 3^2} = \frac{0.8}{10.8} = 0.07407 = 74.07 \times 10^{-3}$$ Rounding to the nearest integer yields 74. ### Pattern Recognition Finding the total moles using the Ideal Gas Law from the final equilibrium pressure and volume cuts down steps, as it avoids explicitly computing the initial hydrogen amount 'a$a$' first. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium