An 8 L8L container holds 5 5 mol and 9 9 mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from 320^oK320oK to 450 ^oK450oK. How much does the pressure change by?

1 Answer
Jun 22, 2016

The pressure decreases by 2316.4" ""kPa"2316.4 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 5 + 9 = 14.

We can write the equation as:

A_((g))+2B_((g))rightarrowAB_(2(g))A(g)+2B(g)AB2(g)

This tells us that 1 mole of AA reacts with 2 moles of BB to give 1 mole of AB_2AB2.

So 9 moles of BB will react with 4.5 moles of AA to give 4.5 moles of AB_2AB2.

This means that there must be an excess 0.5 moles of AA which have not reacted.

So the total moles of gas after reaction = 0.5 + 4.5 = 5.

The ideal gas equation gives us :

PV=nRTPV=nRT

PP is the pressure

VV is the volume

nn is the number of moles

RR is the gas constant with the value 8.31" ""J/K/mol"8.31 J/K/mol

TT is the absolute temperature

Initial:

P_1V=n_1RT_1P1V=n1RT1

:.P_1=(n_1RT_1)/V

P_1=(14xx8.31xx320)/(0.008)=4653600" ""N/m"^2

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

Final:

P_2=(n_2RT_2)/V

:.P_2=(5xx8.31xx450)/0.008=2337187.5" ""N/m"^2

:.DeltaP=P_1-P_2=4653600-2337187.5=2316412.5" ""N/m"^2

DeltaP=2316.4" ""kPa"

(1"N/m"^2=1"Pa")

So you can see that the pressure has dropped by 2316.4" ""kPa"

Although the temperature has increased this is more than offset by the decrease in the number of particles brought about by the reaction, hence the reduction in pressure.