Question

A `5 L` container holds `15` mol and `8` mol of gasses A and B, respectively. Every four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from `250^oK` to `370 ^oK`. By how much does the pressure change?

Answer 1

The final pressure is 1.4 times whatever the initial pressure was.

Explanation:

`3A + 4B → A_3B_4` With an initial ratio of 15:8 gas B is the limiting reagent. Only 8 * 3/4 = 6 moles of A will be combined with 8 moles of B.

Thus, the final composition in the container will be 9 moles of A and 2 moles of `A_3B_4` for a total of 11 moles. The original number of moles was 23.

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 = (11/23)*(370)/(250)`
`P_1/P_2 = 0.71 ; P_2 = P_1 * 1.4` so the final pressure is 1.4 times whatever the initial pressure was (exothermic reaction!).