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?

Question

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?

Nora Arzate · Accepted Answer

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

Before reaction, we have: 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"#: 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"#

Olivia Askew · Answer

undefined To find the pressure change, first, calculate the initial pressure using the ideal gas law $PV = nRT$ where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature. Then, use the combined gas law to find the final pressure with the new temperature. Finally, subtract the initial pressure from the final pressure to find the pressure change. Given that $V = 7 L$, $n_A = 21 mol$, $n_B = 12 mol$, $T_1 = 350 K$, and $T_2 = 420 K$, the initial pressure can be calculated. Then, using the combined gas law, the final pressure can be found. Subtracting the initial pressure from the final pressure gives the pressure change.