How do you solve this chemistry question? Consider the following reaction at 298 K.

Consider the following reaction at 298 K.

2 SO2(g) + O2(g) 2 SO3(g)

An equilibrium mixture contains O2(g) and SO3(g) at partial pressures of 0.50 atm and 2.0 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO2 in the mixture.

In the appendix I have the following:

SO2(g)
ΔG(f)= -300kJ/mol

O2(g)
ΔG(f)= 0kJ/mol

SO3(g)
ΔG(f)= -371kJ/mol

1 Answer
Michael
Mar 19, 2018

You can do it like this:

Explanation:

sf(2SO_(2(g))+O_(2(g))rightleftharpoons2SO_(3(g))

For which:

sf(K_p=(p_(SO_3)^2)/(p_(SO_2)^2xxp_(O_2))

If we find the standard free energy change sf(DeltaG^@) for this reaction, we can find sf(K_p) from which we can get sf(p_(SO_2)).

The relationship between sf(DeltaG^@) and the reaction quotient is given by:

sf(DeltaG=DeltaG^@+RTlnQ)

At equilibrium sf(DeltaG=0). Now the reaction quotient sf(Q) is equal to the equilibrium constant so this becomes:

sf(DeltaG^@=-RTlnK_p)

We can find sf(DeltaG^@) from the data given in appendix 1. We can apply Hess' Law for which:

sf(DeltaG^@) is equal to the total free energy of formation of the products minus the total free energy of formation of the reactants.

sf(DeltaG^@=SigmaDeltaG_(f."prod")-SigmaDeltaG_(f."react"))

sf(DeltaG^@=(2xx-371)-(2xx-300)color(white)(x)kJ)

sf(DeltaG^(@)=-742+600=-142color(white)(x)kJ)

sf(DeltaG^@=-RTlnK_p)

:.sf(lnK_p=(-DeltaG^@)/(RT))

sf(lnK_p=-(-142xx10^3)/(8.31xx298))

sf(lnK_p=57.34)

sf(K_p=8.0xx10^(24))

This is such a large number that we can infer that the position of equilibrium is so far to the right that the equilibrium partial pressure of sf(SO_2) is negligible.

To test this you could put the numbers into the expression for sf(K_p) and solve for sf(p_(SO_2)).

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