A 5 L container holds 8 mol and 5 mol of gasses A and B, respectively. Groups of five molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from 100 K to 240 K. How much does the pressure change?

2 Answers
Dec 26, 2017

The reaction decreases the number of moles of gas in the (constant volume) container from 13 to 7 while increasing the temperature from 100 K to 240 K.

The pressure change is 2794-2162=632 kPa

Explanation:

First we need to write a balanced chemical equation for the reaction:

2A+5B->A_2B_5

Since we have 5 mol of B, all of it will be used up in the reaction. Only 2 of the 8 mol of A will be used, so 6 mol will remain.

After the reaction, therefore, there will be 6 mol of A and 1 mol of B for a total of 7 mol. There were initially 13 mol of gas in the container.

We remember that a mole of any gas that behaves like an ideal gas takes up the same volume under the same temperature and pressure conditions.

Let's call the pressure before the reaction P_1 and after the reaction the pressure is P_2.

PV=nRT where R is the gas constant, 8.314 LkPaK^-1mol^-1

P_1=(nRT)/V=(13xx8.314xx100)/5=2162 kPa

P_2=(nRT)/V=(7xx8.314xx240)/5=2794 kPa

The pressure change is 2794-2162=632 kPa

Dec 27, 2017

29% increase

Explanation:

Idea gas law:

PV=nRT

For mix gases inside the same volume and temperature

PV = (n_A+n_B)RT

Let (n_1, P_1, V_1, T_1) and (n_2, P_2, V_2, T_1) denote the number of moles, pressure, volume, and temperature before and after the reaction happened respectively.

Before the reaction, there are

n_1= n_A+n_B= 8 + 5= 13 mol of mixed gases

The reaction

cancel(5B -> 2A) (This reaction is incorrect, I misread the question, B binds with A, not binds to form A)

5B + 2A -> A_2B_5

Takes away 5 moles of B and 2 moles of A into one moles of A_2B_5, leaving 6 mol of A.

n_2= n_A+n_(A_2B_5) = 6+1 = 7 mol of mixed gases.

The idea gas law before and after the reaction are:

P_1V_1 = n_1 RT_1

P_2V_2 = n_2RT_2

(P_2V_2)/(P_1V_1) = n_2/n_1(RT_2)/(RT_1)

For a fixed container, V_1 = V_2

P_2/(P_1) = 7/13 * (240K)/(100K) = 1.29

The change in pressure is

(P_2-P_1)/p_1 = (DeltaP)/P_1 =P_2/P_1 -1 = 0.29 or 29%

The pressure has thus increased 85%.

Since the volume of the container is given, P_1 can be reality calculated to determine the absolute change in pressure (DeltaP)

P_1 = (n_1 RT_1)/V = ((13mol*8.3145 J/(mol K)*100K)/(5L*0.001m^3/L)) /(1.01325 "x" 10^5 N/(m^2atm)) ~ 21.34 atm

DeltaP ~0.29P_1= 0.29* 21.3 atm = 6.2 atm