Question

Consider the system at 25.0°C A(g) + 2B(g) + C(g) 2D(g) At time zero, only A, B and C are present in a sealed container. The reaction reaches equilibrium 10 minutes after the reaction is initiated. The partial pressures of A, B, C and D are written as Pa, PB, Pc and PD. Pc at 10 min is Pc at 9 min. v Po at 11 min is Po at 12 min. Select Greater than More information required Equal to Less than After the system is at equilibrium, more of gas B is added. After the system returns back to equilibrium, K before the addition of B is K after the addition of B. At equilibrium, Q is K. Equal to PA at 5 min is PA at 7 min. Select K for K for the forward reaction is the reverse reaction. Equal to

Answer 1

Pc is the partial pressure of C gas .

More the mole fraction of gas, more will be the partial pressure.

As the reaction reaches equilibirium after 10 minutes, thus, after that concentration of C remains as constant but before that (i.e., at 9 minutes), the concentration and hence pressure of C is continuously decreasing.

As concentration of C at 10 minutes is less than at 9 minutes.

Thus, Pc at 10 is less than Pc at 9 minutes.

Part 2:

Once the reaction reaches equilibirium, its concentration will remain as constant .Thus, after 10 minutes the concentration and hence pressure will remain as constant .

Thus, Pd at 11 minutes is equal to Pd at 12.

Part 3 :

The k value for a reaction depends upon temperature only and is independent of concentration .

Thus, k below and after is equal .

Part 4 :

You have done it correct.

Answer is equal to

Part 5 :

Before equilibrium, the concentration of reactants is continuously decreasing, thus, concentration of A at 5 minutes is greater than concentration of A at 7 minutes.

Thus, Pa at 5 minutes is greater than Pa at 7 minutes.

Part 6 : you have done it correct .

Answer is equal to.