The equilibrium constants for the reaction, calculated from the partial pressure

Question

The equilibrium constants for the reaction, calculated from the partial pressures of the three hypothetical species in gas phase,

A(g)    +    B(g)    =    2C(g)

are Kp = 137 and 3849 at 500 K and 900 K, respectively.

Which on of the following statements is true?

The reaction is exothermic.
The partial pressure of C(g) is less at 500 K than at 900 K.
Kp is more than Kc by a factor of RT.
The total pressure of the reaction mixture at 500 K is more than that at 900 K.
Higher total pressure shifts the equilibrium to the left

Explanation / Answer

Analysis:

A(g)    +    B(g)    =    2C(g)

Kp = P-C^2 /((P-A)(P-B))

are Kp = 137 and 3849 at 500 K and 900 K, respectively.

The reaction is exothermic. FALSE, since increasing T, which implies endothermic reaction, Kp increases, meaning that more products form vs. reactants

The partial pressure of C(g) is less at 500 K than at 900 K. C is a product, so it will increase in equilibrium, since it is favoured, i.e. C forms whereas A and B react

Kp is more than Kc by a factor of RT. False, Kp and Kc can be related via:

Kp = Kc*(RT)^dn, since dn = mol of products - mol of reactants, then dn = 2-2 = 0, so

Kp = Kc*(RT)^0

Kp = Kc*1

Kp = Kc, no effect

The total pressure of the reaction mixture at 500 K is more than that at 900 K. False, the pressure is the same, since 2 mol of A + B reac tto form 2 mol of BC

Higher total pressure shifts the equilibrium to the left. Since 2 mol of gas ar epresent in both sides of the reaction, pressurizaiton will not shift equilibrium

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