A 7 L container holds 7 mol and 18 mol of gasses A and B, respectively. Every three of molecules of gas B bind to one molecule of gas A and the reaction changes the temperature from 320^oK to 240 ^oK. By how much does the pressure change?

1 Answer
Aug 25, 2016

The pressure decreases by sf(1462color(white)(x)kPa)

Explanation:

The key to this problem is to find the number of moles of gas before and after the reaction.

We can then use the ideal gas equation to find the pressure in each case and hence the pressure change.

Before the gases react we have a total number of moles of 7 + 18 = 25

We can write the equation as:

sf(3B_((g))+A_((g))rarrAB_(3(g))

This tells us that 1 mole of sf(A) reacts with 3 moles of sf(B) to give 1 mole of sf(AB_3).

Multiplying through by 6 tells us that 18 moles of sf(B) will react with 6 moles of sf(A) to give 6 moles of sf(AB_3).

Since we have 7 moles of sf(A) we have 7 - 6 = 1 mole of sf(A) which is in excess.

So the total moles after reaction = 6 + 1 = 7.

:. Total initial moles = 25

Total final moles = 7

The ideal gas equation gives us :

sf(PV=nRT)

sf(P) is the pressure

sf(V) is the volume

sf(n) is the number of moles

sf(R) is the gas constant with the value sf(8.31color(white)(x)"J/K/mol")

sf(T) is the absolute temperature

Initial:

sf(P_1V=n_1RT_1)

:.sf(P_1=(n_1RT_1)/V)

sf(P_1=(25xx8.31xx320)/(0.007)=1.662xx10^(7)color(white)(x)"N/m"^2)

(note I have converted sf("L") to sf("m"^3))

Final:

sf(P_2=(n_2RT_2)/V)

sf(P_2=(7xx8.31xx240)/(0.007)=0.19944xx10^(7)color(white)(x)"N""/""m"^2)

:.sf(DeltaP=(1.662-0.19944)xx10^(7)=1.462xx10^(7)color(white)(x)"N""/""m"^2)

So the pressure has been reduced by sf(1462color(white)(x)kPa)