A 7 L7L container holds 38 38 mol and 15 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 140^oK140oK to 210^oK210oK. How much does the pressure change?

1 Answer
Feb 21, 2017

The final pressure is 0.93 times whatever the initial pressure was.

Explanation:

5A + 3B → A_5B_35A+3BA5B3 With an initial ratio of 38:15 gas B is the limiting reagent. Only 15 * 3/5 = 9.0 moles of A will be combined with 15 moles of B.

Thus, the final composition in the container will be 29 moles of A and 9 moles of A_5B_3A5B3 for a total of 38 moles. The original number of moles was 53.

Assuming ideal gas behavior where n = PV/RT , with a constant volume , the ratio of the temperature change in .^oK.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(n1PVT)1=(n2PVT)2 ; (n_1/n_2)*(T_1/T_2) = P_2/P_1(n1n2)(T1T2)=P2P1

P_1/P_2 = (n_2/n_1)*(T_2/T_1)P1P2=(n2n1)(T2T1)

P_1/P_2 = (38/53)*(210)/(140)P1P2=(3853)210140

P_1/P_2 = 1.07 ; P_2 = P_1 * 0.93P1P2=1.07;P2=P10.93
so the final pressure is 0.93 times whatever the initial pressure was.