A 7L container holds 38 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 140oK to 210oK. 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+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 A5B3 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 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.

(n1PVT)1=(n2PVT)2 ; (n1n2)(T1T2)=P2P1

P1P2=(n2n1)(T2T1)

P1P2=(3853)210140

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