A 7 L container holds 21 mol and 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^oK to 420^oK. How much does the pressure change?

1 Answer
Miguel F.
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"
  • n_"A1" = 21 " mol", n_"B1" = 12 " mol"
  • T_1 = 350 " K"

Applying the law:

P_1 V_1 = n_1 R T_1 rightarrow

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

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

Now, let us write the reaction: 3 molecules of "B" + 5 molecules of "A" form "A"_5 "B"_3.

5 "A" + 3 "B" rightarrow "A"_5 "B"_3

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

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

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

Thus, reaction is:

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

So we'll have in the end: n_2 = n_{"A"5"B"3} + n_"B" = 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"

And the change on pressure is:

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

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