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?

Answer 1

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"`