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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