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For the reaction NO(g) + 1/2 O2(g) -----> at 750 degrees Celsius, the equilibrium constant Kc equals

For the reaction NO(g) + 1/2 O2(g) -----> at 750 degrees Celsius, the equilibrium-example-1
User Bo Qiang
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Step-by-step explanation:

Given the following reaction.

NO (g) + 1/2 O₂ (g) <---> NO₂ (g)

We have to find the relationship between its Kc and Kp.

The Kp can be calculated from the Kc using the following expression.

Kp = Kc * (R*T)^(Δn)

Where Δn is the sum of the coefficients of the products minus the sum of the coefficients of the reactants. So Δn will be equal to:

Δn = Σ n of products - Σ n of reactants

Δn = 1 - (1 + 1/2)

Δn = -1/2

So we can say that:

Kp = Kc * (R*T)^(Δn)

Kp = Kc * (R*T)^(-1/2)

Kc = Kp * (R*T)^(1/2)

Answer: Kc = Kp * (R*T)^(1/2)

User Benny Hill
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