The equilibrium constant K_c, is calculated using molar concentrations For gaseo

Question

The equilibrium constant K_c, is calculated using molar concentrations For gaseous reactions another form of the equilibrium constant. K_p is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation K_p = K_c(RT)^delta n. where R = 0.08206 L middot atm/(K middot mol). T is the absolute temperature, and delta n is the change in the number of moles of gas (sum moles products - sum moles reactants) For example consider the reaction N_2(g) + 3H_2(g) = 2NH_3(g) for which delta n = 2 - (1 + 3) = - 2 For the reaction 3A(g) + 2B(g) C(g) K_c = 92.2 at a temperature of 101 degree C Calculate the value of K_p. Express your answer numerically. For the reaction X(g) + 3Y(g) 3Z(g) K_c = 2.08 times 10^-2 at a temperature of 35 degree C. calculate the value of K_c. express your answer numerically.

Explanation / Answer

Part A:

Kc=92.2, T=101°C=101+273.16=374.16 K, R=0.08206 L.atm/(k.mol),

n=Sum of the moles of the products - Sum of the moles of reactants=1-(3+2)=-4

Kp=Kc(RT)n=92.2 (0.08206 X 374.16)-4=92.2 X 0.00000112523=0.00010374697=1.04 X 10-4

Part B:

Kp=2.08 X 10-2, T=35°C=35+273.16=308.16 K, R=0.08206 L.atm/(k.mol),

n=Sum of the moles of the products - Sum of the moles of reactants=3-(1+3)=-1

Kc=Kp / (RT)n=2.08 X 10-2 / (0.08206 X 308.16)-1=0.0208 / 0.03954506=0.52598=5.26 X 10-1

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