Question

Carbon monoxide and molecular hydrogen react to form methanol. A 5.70 L reactor at 25.0 oC is charged with 2.56 bar of CO. The gas is then pressurized with H2 to give a total pressure of 3.26 bar. The reactor is sealed, heated to 350 o C to drive the reaction to completion, and cooled back to 25.0 o C. Compute the final partial pressure of each gas (in bar) after the reaction has occurred.

Answer 1

The balanced equation for the reaction between CO and H2 can be written as follows.

CO(g) + 2H₂(g) CH₃OH ............. Equation 1

The volume of the reactor (V) = 5.7 L

Temperature (T) = 25 + 273.15 = 298.15 K

The pressure of CO (P_CO) = 2.56 bar

The pressure of H2 (P_H2) = 3.26-2.56 = 0.7 bar

Universal gas constant (R) = 0.08314 L bar mol^-1 K^-1

According to ideal gas equation: PV = nRT

i.e. (n_CO)initial = 2.56 bar * 5.7 L/(0.08314 L bar mol^-1 K^-1 * 298.15 K) = 0.589 mol

(n_H2)initial = 0.7 bar * 5.7 L/(0.08314 L bar mol^-1 K^-1 * 298.15 K) = 0.161 mol

Here, H2 is the limiting reactant, i.e. the no. of moles of H2 after reaction (n_H2)final = 0 mol

Therefore, the no. of moles of CO remaining (n_CO)final = 0.589 - (0.161/2) = 0.508 mol

Hence, according to the ideal gas equation, the final partial pressure of CO

(P_CO)final = (n_CO)final * R * T/V = 0.508 mol * 0.08314 L bar mol^-1 K^-1 * 298.15 K/5.7 L = 2.21 bar

And (P_H2)final = 0 bar

(P_CH3OH)final = (n_CH3OH) * R * T/V = 0.0805 mol * 0.08314 L bar mol^-1 K^-1 * 298.15 K/5.7 L = 0.35 bar