Question

A `3 L` container holds `30` mol and `15` mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from `320^oK` to `480^oK`. How much does the pressure change?

Answer 1

The final pressure is three times whatever the initial pressure was.

Explanation:

`5A + 3B → A_5B_3` With an initial ratio of 2:1 gas B is the limiting reagent. Only 15 * 5/3 = 25 moles of A will be combined with 15 moles of B.

Thus, the final composition in the container will be 5 moles of A and 5 moles of `A_5B_3` for a total of 10 moles. The original number of moles was 45.

Assuming ideal gas behavior where n = PV/RT , with a constant volume, the ratio of the temperature change in oK is the same as the ratio of the change in the total number of moles of gas in the container times the inverse ratio of the pressures.

`(n_1*(PV)/T)_1 = (n_2*(PV)/T)_2` ; `(n_1/n_2)*(T_1/T_2) = P_2/P_1`
`P_1/P_2 = (n_2/n_1)*(T_2/T_1)`
`P_1/P_2 = (10/45)*(480)/(320)`
`P_1/P_2 = 0.3333 ; P_2 = P_1 * 3` so the final pressure is three times whatever the initial pressure was (exothermic reaction!).