Question

write the pressure equilibrium constant expression for this reaction.
2ch₃oh(l)+3o₂(g)→2co₂(g)+4h₂o(g)

Explanation:

Step1: Recall the formula for pressure equilibrium constant

For a reaction \(aA + bB
ightleftharpoons cC + dD\) (where \(A,B,C,D\) are gases), \(K_p=\frac{P_C^c\times P_D^d}{P_A^a\times P_B^b}\). Liquids are excluded from the expression.

Step2: Identify the gases in the reaction

In \(2CH_3OH(l)+3O_2(g)
ightleftharpoons2CO_2(g)+4H_2O(g)\), \(CH_3OH\) is liquid (\(l\)), \(O_2\), \(CO_2\) and \(H_2O\) are gases (\(g\)).

Step3: Substitute into the formula

Here \(a = 3\) (\(O_2\)), \(c = 2\) (\(CO_2\)), \(d=4\) (\(H_2O\)). So \(K_p=\frac{P_{CO_2}^2\times P_{H_2O}^4}{P_{O_2}^3}\)

Answer:

\(K_p=\frac{P_{CO_2}^2\times P_{H_2O}^4}{P_{O_2}^3}\)