The equilibrium constant, Kc, is calculated using molar concentrations. For gase

Question

The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)n where R=0.08206 Latm/(Kmol), T is the absolute temperature, and n is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction N2(g)+3H2(g)2NH3(g) for which n=2(1+3)=2.

A) For the reaction 3A(g)+3B(g)C(g) Kc = 41.6 at a temperature of 23 C . Calculate the value of Kp. Express your answer numerically.

B)For the reaction X(g)+3Y(g)2Z(g) Kp = 3.18×102 at a temperature of 149 C . Calculate the value of Kc. Express your answer numerically.

Explanation / Answer

A) 3A(g)+3B(g)C(g)

Kc = 41.6 at a temperature of 23 C

n=1(3+3)=5.

T = 273 + 23 = 296 K

Hence Kp = 41.6x (0.08206 x296)^-5

Kp = 4.92*10^-6

B)

X(g)+3Y(g)2Z(g)

Kp = 3.18×102 at a temperature of 149 C

Calculate the value of Kc.
Here, n = 2 (1+3) = 2. and T = 273 + 149= 422 K

3.18×102 = Kc x (0.08206 x422)^-2

Thus Kc = (3.18×10^2)*(0.08206 x422)^2

Kc = 38.134

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