ou are given the following data.

S(s) + 3/2 O2(g) → SO3(g) ΔH = -395.2 kJ
2 SO2(g) + O2(g) → 2 SO3(g) ΔH = -198.2 kJ

Calculate ΔH for the reaction below.
S(s) + O2(g) → SO2(g)

delta H=-395.2kJ+1/2 198.2

To find the enthalpy change (ΔH) for the reaction: S(s) + O2(g) → SO2(g), we can use the given enthalpy changes for the other reactions and manipulate them to match the desired equation.

Steps:
1. Start by reversing the first equation: SO3(g) → S(s) + 3/2 O2(g). (Since we want the S(s) term to be on the product side)
ΔH = +395.2 kJ (note the change in sign since we are reversing the equation)

2. Multiply the second equation by 2: 2 SO2(g) + 2 O2(g) → 2 SO3(g).
ΔH = -198.2 kJ (no change in sign)

3. Now we need to manipulate the equations to match the desired reaction:
a) Multiply the first equation by 2, so the terms can be balanced with the second equation:
2 SO3(g) → 2 S(s) + 3 O2(g)
ΔH = 2*(-395.2 kJ) = -790.4 kJ

b) Reverse the second equation:
-2 SO3(g) → -2 SO2(g) - 2 O2(g)
ΔH = -2*(-198.2 kJ) = 396.4 kJ (note the change in sign)

4. Add the manipulated equations to get the desired reaction equation:
-2 SO3(g) + 2 SO2(g) + O2(g) = -2 SO2(g) - 2 O2(g) + 2 S(s) + 3 O2(g)

5. Simplify the equation by canceling out identical terms on both sides:
-2 SO3(g) + 2 SO2(g) = -2 SO2(g) + 2 S(s)

6. The resulting equation is:
-2 SO3(g) + 2 SO2(g) = -2 SO2(g) + 2 S(s)
ΔH = ΔH1 + ΔH2 = -790.4 kJ + 396.4 kJ
ΔH = -394 kJ

Therefore, the enthalpy change (ΔH) for the reaction: S(s) + O2(g) → SO2(g) is -394 kJ.

To calculate the enthalpy change (ΔH) for the reaction S(s) + O2(g) → SO2(g), we can use Hess's Law. Hess's Law states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction.

Here's how you can calculate ΔH for the given reaction:

1. Start by looking at the given reactions and their corresponding enthalpy changes:
a) S(s) + 3/2 O2(g) → SO3(g) ΔH = -395.2 kJ
b) 2 SO2(g) + O2(g) → 2 SO3(g) ΔH = -198.2 kJ

2. In order to make the second reaction match the desired reaction, we need to reverse it and multiply it by a factor of 1/2. This will give us the desired products of S(s) + O2(g) → SO2(g).
c) 2 SO3(g) → 2 SO2(g) + O2(g) ΔH = +198.2 kJ

3. Now, we can add the two reactions together to cancel out the common compounds:
a) S(s) + 3/2 O2(g) → SO3(g) ΔH = -395.2 kJ
c) 2 SO3(g) → 2 SO2(g) + O2(g) ΔH = +198.2 kJ

Adding a + b gives:
S(s) + 2 SO3(g) → SO2(g) + 3/2 O2(g) ΔH = -197 kJ

4. Finally, since we want the reaction S(s) + O2(g) → SO2(g), we can multiply the equation by a factor of 1/2 to get the final balanced equation and the corresponding ΔH:
1/2 * (S(s) + 2 SO3(g) → SO2(g) + 3/2 O2(g)) = 1/2 * (-197 kJ)

The final balanced equation and ΔH for the reaction S(s) + O2(g) → SO2(g) would be:
S(s) + 2/2 SO3(g) → 1/2 SO2(g) + 3/4 O2(g) ΔH = -98.5 kJ

Therefore, the enthalpy change (ΔH) for the reaction S(s) + O2(g) → SO2(g) is -98.5 kJ.