Question

2. A 2.00 L flask containing 0.600 atm of gas A is connected to a 4.00 L flask containing 0.600 atm of gas B, after which the valve between the flasks is opened so that the gases can mix and react according to the following reaction equation. If the theoretical yield is produced at constant temperature, the total pressure in the combined flasks should be....? 2A(g) + 3B(g) → A2B3(g)

Answer 1

0.500 atm

Step-by-step explanation:

For the reaction

`2A(g) +3B(g) - -> A_(2)B_(3)(g)`

Assuming gas ideal

PV=nRT

P= pressure, V= volume of the container, n= mol of gas, R=constant of gases and T=temperature.

mol of A =
`(P_(A)V_(A))/(RT)`

mol of B=
`(P_(B)V_(B))/(RT)`

we don't know the temperature but we know that is constant so is the same in both flasks.

we need to know if one of the reagents is limiting the reaction or if they are in right proportions. When we see the expressions for the moles for each reagent we see that both have RT in the denominator and this value is constant so we can say that the number of moles of each reagent is proportional to just PiVi

`P_(A)V_(A)=0.600 atm *2.00 L= 1.2atm*L`

`P_(B)V_(B)=0.600 atm *4.00 L= 2.4atm*L`

the stoichiometry relationship is:

3 mol of B need 2 mol of A

Applying the proportionality

`3 P_(B)V_(B) : 2P_(A)V_(A)`

so 2.4 atm*L of B need
`(2)/(3)*2.4atm*L= 1.6 atm*L` of A

we just have 1.2atm*L of A so, this is the limiting reagent of the reaction.

Assuming theoretical yield we can say that all A was consumed and in the final condition we have the B that did not reacted and the product A2B3

The dalton's law tell us that the total pressure of a mixture of gases is the sum of the partial pressure of every gas.

So after the reaction the total pressure is:

`P_(total)=P_(A2B3) + P_(B)`

we can also say that the volume in the final condition is

`V_(2)=V_(A)+V_(B)`

`P_{{A_(2)B_(3)}=\frac{n_{A_(2)B_(3)}*R*T}{V_(2)}`

`P_(B)=(n_(B)*R*T)/(V_(2))`

so P total is

`P_(total)= (n_{A_(2)B_(3)}+n_(B))*(R*T)/(V_(2))`

From the reaction stoichiometry we know that 2 mol of A produce 1 mol of
`{A_(2)B_(3)`

so

`n_{A_(2)B_(3)}=2*(P_(A)V_(A))/(RT)`

moles of B in the outlet would be

`nB_(out)=nB_(in)-nB_(used)`

the moles of B used or consumed in the reaction are

mol of B = (3/2)A

`nB_(out)=nB_(in)-(3)/(2)*nA_(in)`

`nB_(out)=(P_(B)V_(B))/(RT)-(3)/(2)*(P_(A)V_(A))/(RT)`

replacing
`n{{A_(2)B_(3)}` and nB in the total pressure equation we get:

`P_(total)= (2*(P_(A)V_(A))/(RT)+(P_(B)V_(B))/(RT)-(3)/(2)*(P_(A)V_(A))/(RT))*(R*T)/(V_(2))`

`P_(total)=(P_(B)V_(B) +0.5*P_(A)V_(A))/(V_(2))`

`P_(total)= ((0.600 atm *4.00 L)+0.5*0.600 atm *2.00 L)/(6L)=0.500atm`