Question

C2H4(9) + H20(g) = CH:CH20H(g), 4Н= - 46 kJ/mol Write the expression for the equilibrium constant in terms of pressure, Kp, for this reaction.

Answer 1

The expression for the equilibrium constant in terms of pressure, Kp, for the reaction C2H4(g) + H2O(g) = CH3CH2OH(g), is Kp = P_CH3CH2OH / (P_C2H4 × P_H2O).

Step-by-step explanation:

To write the expression for the equilibrium constant in terms of pressure, Kp, for the reaction:
C2H4(g) + H2O(g) = CH3CH2OH(g), we use the partial pressures of the gases. The equation given is ΔH = -46 kJ/mol, but this information is not needed for the Kp expression itself. The general form for the equilibrium constant expression for a reaction aA + bB = cC + dD is given by:

Kp = (PCc × PDd) / (PAa × PBb)

Where P represents the partial pressure of the gas and the lowercase letters represent the stoichiometric coefficients from the balanced equation. Applying this formula to our reaction we get:

Kp = PCH3CH2OH / (PC2H4 × PH2O)

This represents the equilibrium pressure relationship for the provided reaction. In this case, there is no need to account for changes in the number of moles of gas since the reaction involves equal amounts of reactant and product gases (1 mole each).