A 55 LL container holds 8 8 mol and 5 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 100100 KK to 240240 KK. 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 1313 to 77 while increasing the temperature from 100100 KK to 240240 KK.

The pressure change is 2794-2162=63227942162=632 kPakPa

Explanation:

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

2A+5B->A_2B_52A+5BA2B5

Since we have 55 mol of BB, all of it will be used up in the reaction. Only 22 of the 88 mol of AA will be used, so 66 mol will remain.

After the reaction, therefore, there will be 66 mol of AA and 11 mol of BB for a total of 77 mol. There were initially 1313 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_1P1 and after the reaction the pressure is P_2P2.

PV=nRTPV=nRT where RR is the gas constant, 8.3148.314 LkPaK^-1mol^-1LkPaK1mol1

P_1=(nRT)/V=(13xx8.314xx100)/5=2162P1=nRTV=13×8.314×1005=2162 kPakPa

P_2=(nRT)/V=(7xx8.314xx240)/5=2794P2=nRTV=7×8.314×2405=2794 kPakPa

The pressure change is 2794-2162=63227942162=632 kPakPa

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=(nA+nB)RT

Let (n_1, P_1, V_1, T_1) and (n_2, P_2, V_2, T_1) (n1,P1,V1,T1)and(n2,P2,V2,T1) 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 n1=nA+nB=8+5=13mol 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