Given: C(s)+ CO₂(g) → 2CO(g) At equilibrium at a certain temperature,the partial pressures of CO(g)and CO₂(g)are 1.44 atm and 0.820 atm,respectively.Calculate the value of K for this reaction.

Question

Quizplus · Accepted Answer

The equilibrium constant (K) for a reaction is calculated using the partial pressures of the reactants and products at equilibrium. In this case, the partial pressure of CO(g) is 1.44 atm and the partial pressure of CO2(g) is 0.820 atm. The reaction is:C(s) + CO2(g) → 2CO(g)The equilibrium constant expression is:K = (PCO)^2 / PCO2Substituting the partial pressures into the expression, we get:K = (1.44 atm)^2 / 0.820 atm = 2.53Therefore, the value of K for this reaction is 2.53.

Given: C(s)+ CO2(g) → 2CO(g) at Equilibrium at a Certain

Given: C(s)+ CO₂(g) → 2CO(g) at Equilibrium at a Certain