A 7 L7L container holds 21 21 mol and 12 12 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 350^oK350oK to 420^oK420oK. How much does the pressure change?

1 Answer
Jul 3, 2016

Let us assume that both gases are ideal gases, so we may use the ideal gas law.

Explanation:

Before reaction, we have:

  • V_1 = 7 " L"V1=7 L
  • n_"A1" = 21 " mol"nA1=21 mol, n_"B1" = 12 " mol"nB1=12 mol
  • T_1 = 350 " K"T1=350 K

Applying the law:

P_1 V_1 = n_1 R T_1 rightarrow P1V1=n1RT1

rightarrow P_1 = {n_1 R T_1}/{V_1} = {39 " mol" cdot 0,082 " atm·L/K·mol" cdot 350 " K"}/{7 " L"}P1=n1RT1V1=39 mol0,082 atm⋅L/K⋅mol350 K7 L
= 159.9 " atm"=159.9 atm

Where we have used that n_1 = n_"A1" + n_"B1" = 21 + 12 " mol"n1=nA1+nB1=21+12 mol.

Now, let us write the reaction: 3 molecules of "B"B + 5 molecules of "A"A form "A"_5 "B"_3A5B3.

5 "A" + 3 "B" rightarrow "A"_5 "B"_35A+3BA5B3

With our initial conditions, taking in account that we need 5/3 = 1.6753=1.67 times more "A"A than "B"B:

  1. We would need, for 21 mol of "A"A, 3/5 cdot 21 = 12.63521=12.6 mol of "B"B.
  2. We would need, for 12 mol of "B"B, 5/3 cdot 12 = 205312=20 mol of "A"A.

Given that we just have 12 mol of "B"B, the option 1 is not possible, so we choose option 2 ("A"A is limiting reactant, and "B"B is excess reactant).

Thus, reaction is:

20 "A" + 12 "B" rightarrow 4 "A"_5 "B"_3 + 1 "B"20A+12B4A5B3+1B

So we'll have in the end: n_2 = n_{"A"5"B"3} + n_"B" = 4 + 1 = 5 " mol"n2=nA5B3+nB=4+1=5 mol.

By applying ideal gas law again:

P_2 = {5 " mol" cdot 0,082 " atm·L/K·mol" cdot 420 " K"}/{7 " L"} = 24.6 " atm"P2=5 mol0,082 atm⋅L/K⋅mol420 K7 L=24.6 atm

And the change on pressure is:

Delta P = P_2 - P_1 = 24.6 - 159.9 = -135.3 " atm"