Use the equations with the enthalpy information given below to calculate the ÄH° for the reaction:

1) S(s) + 3/2 O2(g) --> SO3(g)

ÄH° = -395kJ

2) 2SO2(g) + O2(g) --> 2 SO3(g)
ÄH° = -198.2 kJ

Multiply equation 1 by 2 and add to the reverse of equation 2. delta H values are multiplied by any multipliers and delta H values change their sign when the reaction is reversed.
When you finish you will have the right equation EXCEPT it will be just twice what you want; therefore, divide everything by 2 including the delta H value.

How much heat is evolved when 320 g of SO2 is burned according to the chemical equation shown below?

To find the ΔH° for the reaction S(s) + O2(g) → SO2(g), we can use the given equations and their enthalpy values.

Step 1: Multiply the first equation by 2 and reverse it:
2 SO2(g) → 2 S(s) + 2 O2(g)
ΔH° = 2 * (-198.2 kJ) = -396.4 kJ

Step 2: Multiply the second equation by 2:
2 SO2(g) + O2(g) → 2 SO3(g)
ΔH° = 2 * (-198.2 kJ) = -396.4 kJ

Step 3: Add the two modified equations to get the desired reaction:
2 SO2(g) → 2 S(s) + 2 O2(g)
2 SO2(g) + O2(g) → 2 SO3(g)
---------------------------
S(s) + O2(g) → SO2(g)

By adding the equations and their respective enthalpies, we get:
(S(s) + O2(g) → SO2(g)) = (S(s) + O2(g) → SO2(g)) + (2 SO2(g) + O2(g) → 2 SO3(g))
ΔH° = -396.4 kJ + (-396.4 kJ) = -792.8 kJ

Therefore, the ΔH° for the reaction S(s) + O2(g) → SO2(g) is -792.8 kJ.

To calculate the ΔH° for the reaction S(s) + O2(g) → SO2(g), we can use the given enthalpy information for the other two reactions and manipulate them to match the desired reaction.

First, let's reverse the first reaction and change the sign of its enthalpy value:
SO2(g) → S(s) + O2(g)
ΔH° = +395 kJ

Since we want the equation S(s) + O2(g) → SO2(g), we need to add these two reactions together. To do that, we can multiply the first reaction by 2 so that the moles of SO2 cancel out:

2(SO2(g) → S(s) + O2(g)) (1st reaction * 2)
+ (2SO2(g) + O2(g) → 2SO3(g)) (3rd reaction)
_______________________________________
2S(s) + 2O2(g) → 2SO2(g) + 2SO3(g)

Now, we can sum up the enthalpy values of the two reactions:

(2 times the enthalpy of the 1st reaction) + (enthalpy of the 3rd reaction) = ΔH°

2 × (+395 kJ) + (-198.2 kJ) = ΔH°

790 kJ - 198.2 kJ = ΔH°
591.8 kJ = ΔH°

Therefore, the ΔH° for the reaction S(s) + O2(g) → SO2(g) is 591.8 kJ.