The reaction NO_2(g) + NO(g) = N_2O(g) + O_2(5) has Delta G degree_1273 = -9.67

Question

The reaction NO_2(g) + NO(g) = N_2O(g) + O_2(5) has Delta G degree_1273 = -9.67 Id. A 1.00 L reaction vessel at 1000 degree C contains 0.0152 mol NO_2, 0.040 mol NO, 0.018 mol N_2O, and 0.0350 mol O_2. Calculate the values of the equilibrium constant and the reaction quotient. Is the reaction at equilibrium? If not, in which direction will the reaction proceed to reach equilibrium? There is not enough information to decide. The reaction is not in equilibrium and must shift to the right to reach equilibrium. The reaction is not in equilibrium and must shift to the left to reach equilibrium. The system is in equilibrium. Consider the reaction shown below. PbCO_3(s) = PbO(s) + CO_2(g) Calculate the equilibrium pressure of CO_2 in the system at the following temperatures. 110 degree C 510 degree C

Explanation / Answer

Ans 1 ) K = (N20)*(O2) / (NO2) * (NO)

As the reaction is for 1 litre of solution the moles will remain same

K = (0.018) * (0.0350) / (0.0152) * (0.040)

K = 0.00063 / 0.000608

K = 1.03

Q =  (N20)*(O2) / (NO2) * (NO)

Q = (0.018) * (0.0350) / (0.0152) * (0.040)

Q = 0.00063 / 0.000608

Q = 1.03

As Q=K the reaction is in equilibrium

Ans 2 ) To find Delta G at 110 degree i.e 383 k & Delta G at 510 degree i.e 783 k

Del G = del H - T * del S

Del G @ 110 = 88300 j / mol - 383 k * 151.3 j / mol k

Del G @ 110 = 88300 j / mol - 57947 j / mol

Del G @ 110 = 30352 j / mol

Now Del G = - RT lnKp

30352 j / mol = - 8.314 j / mol k * 383 k * lnKp

ln Kp = - 30352 / 3184.2

    ln Kp = - 9.53

Kp = 7.26 * 10^ -5 atm

Similarly

Del G = - RT lnKp

30352 j / mol = - 8.314 j / mol k * 783 k * lnKp

ln Kp = - 30352 / 6509.8

    ln Kp = - 4.66

Kp = 9.46 * 10^ -3 atm

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