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?

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"#