Use the thermochemical equations shown below to determine

the enthalpy (kJ) for the reaction:

H2SO3(l)=>H2S(g) + 3/2O2(g)

H2SO3(l)=>H2O(l) +SO2(g) DH=62KJ

SO2(g)=>S(s) + O2(g) DH=297KJ

H2S(g) + 1/2O2(g)=>S(s) + H2O(l) DH=-155KJ

To determine the enthalpy (ΔH) for the reaction H2SO3(l) => H2S(g) + 3/2O2(g), you can use the given thermochemical equations and apply Hess's Law, which states that the total enthalpy change for a reaction is independent of the pathway taken.

The objective is to manipulate the given thermochemical equations in a way that will cancel out/rearrange the desired reaction, so that the enthalpy values can be added or subtracted to calculate ΔH for the desired reaction.

1. Given equation: H2SO3(l) => H2O(l) + SO2(g) ΔH = 62 kJ

2. Given equation: SO2(g) => S(s) + O2(g) ΔH = 297 kJ

3. Given equation: H2S(g) + 1/2O2(g) => S(s) + H2O(l) ΔH = -155 kJ

Now, let's begin manipulating the equations to obtain the desired reaction.

- We want to obtain H2S(g) on the right side of the equation, so we need to reverse equation 3:
S(s) + H2O(l) => H2S(g) + 1/2O2(g) ΔH = +155 kJ

- We also want 3/2O2(g) on the right side, but equation 1 only provides 1/2O2(g). To get 3/2O2(g) we can multiply equation 1 by 3, resulting in:
3H2SO3(l) => 3H2O(l) + 3SO2(g) ΔH = 3(62 kJ) = 186 kJ

- Lastly, we want S(s) on the left side, but equation 2 only provides S(s) on the right side. Reversing equation 2 gives:
S(s) + O2(g) => SO2(g) ΔH = -297 kJ

Now we can add the manipulated equations together to cancel out the intermediates and obtain the desired reaction:

3H2SO3(l) => 3H2O(l) + 3SO2(g) (multiplied by 186 kJ)
S(s) + O2(g) => SO2(g) (reversed, multiplied by -297 kJ)
S(s) + H2O(l) => H2S(g) + 1/2O2(g) (reversed, multiplied by +155 kJ)

Adding up the equations:
3H2SO3(l) + S(s) + O2(g) + S(s) + H2O(l) => 3H2O(l) + 3SO2(g) + SO2(g) + H2S(g) + 1/2O2(g)

Simplifying:
3H2SO3(l) => H2O(l) + 3SO2(g) + H2S(g)

Since this reaction equation matches our desired equation:

ΔH for the reaction H2SO3(l) => H2S(g) + 3/2O2(g) = 186 kJ + (-297 kJ) + (+155 kJ)
ΔH = 44 kJ

Therefore, the enthalpy change for the reaction H2SO3(l) => H2S(g) + 3/2O2(g) is 44 kJ.