Given the equilibrium constants for the following reactions: 4Cu(s)+O2(g) 2Cu2O(

Question

Given the equilibrium constants for the following reactions: 4Cu(s)+O2(g) 2Cu2O(s) K1 2CuO(s) Cu2O(s) Cu2O(s)+1/2O2(g),K2 what is K for the system 2Cu(s)+O2(g) 2CuO(s) equivelnt to ? For the gaseous reaction, 2 H2 + 2 NO 2 H2O + N2, Kp at 120 degree C = 2.42. At a given moment, it is found that the partial pressures of H2, NO, H20 and N2 are 1.1, 1.3, 0.78 and 2.2 atm, respectively. Which of the following statements describes the situation? The reaction is at equilibrium Qp = 0.65 so the reaction goes to the left Qp = 0.65 so the reaction goes to the right Qp = 1 -2 so the reaction goes to the left

Explanation / Answer

(9)

The final reaction is obtained by the following steps :

Step 1 : Divide reaction by 2 : K1 changes to K11/2

Step 2 : Reverse reaction 2 : K2 changes to 1/K2

Step 3 : Add reactions obtained from steps 1 and 2

When we add the reactions, their K values get multiplied. Thus,

Resultant K = K11/2/K2 = sqrt(K1)/K2

(10)

Given reaction :

2H2 + 2NO ----> 2H2O + N2

Kp = (pH2O2*pN2)/(pH22*pNO2)

Putting the given values of partial pressures to calculate the reaction quotient, Q, we get :

Q = (0.782*2.2)/(1.12*1.32) = 0.65

Kp = 2.42

Since Q < Kp, it means that reaction will proceed to the right in order to attain equilibrium.

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