A 5 L container holds 5 mol and 10 mol of gasses A and B, respectively. Every five of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from 320^oK to 450 ^oK. By how much does the pressure change?

1 Answer
Feb 24, 2018

There is a change of -56.649 atmospheres.

Explanation:

To solve for DeltaP, we need two values: P_"initial" and P_"final".

First, we calculate the initial pressure using the Ideal Gas Law:

P_iV=n_iRT_i

Here,

V=5"L"

n_i=15"mol"

T_i=320"K"

R=0.0821"L atm K"^-1"mol"^_1

To solve for P_i, we rearrange:

P_i=(n_iRT_i)/V

And input:

P_i=(15*0.0821*320)/5

P_i=78.816"atm"

Now, we need the equation that took place. It is:

5"A"+10"B"rarr2"A"_2"B"_5+"A". Here, one mole of "A" is left unreacted.

In the products side, we find we now have 3 moles of gas. Use the Ideal Gas Law again:

P_fV=n_fRT_f

Here,

n_f=3"mol"

T_f=450"K"

Rearranging to solve for P_f:

P_f=(n_fRT_f)/V and inputting:

P_f=(3*0.0821*450)/5

P_f=22.167"atm"

We know that DeltaP=P_f-P_i. Inputting:

DeltaP=22.167-78.816

DeltaP=-56.549"atm"