Enthalpy Change

Given the following reactions:
S(s) + O2(g) → SO2(g) ΔH = -297 kJ/mol
SO2(g) + ½ O2(g) → SO3(g) ΔH = -141 kJ/mol

Calculate the ΔH for the reaction:
S(s) + 3/2O2(g) → SO3(g)

ΔH= (-297)+ (-141)

-438kJ/mol

Well, calculating the ΔH for this reaction is quite simple. All we need to do is add up the ΔH values of the individual reactions.

First, let's double the second reaction to align the coefficients:

2SO2(g) + O2(g) → 2SO3(g) ΔH = -282 kJ/mol

Now, let's multiply the first reaction by 2/2 to align the coefficients:

2S(s) + 2O2(g) → 2SO2(g) ΔH = -594 kJ/mol

Finally, we can combine the two reactions:

2S(s) + 2O2(g) + O2(g) → 2SO2(g) + O2(g) → 2SO3(g) ΔH = -594 kJ/mol + (-282 kJ/mol)

Performing the addition:

ΔH = -876 kJ/mol

So, the ΔH for the reaction S(s) + 3/2O2(g) → SO3(g) is -876 kJ/mol.

And hey, wasn't that reaction just a series of reactions entangled in a complicated dance? Just like trying to follow a complex dance routine while wearing oversized clown shoes!

To calculate the enthalpy change (ΔH) for the reaction:

S(s) + 3/2O2(g) → SO3(g)

we can use the given reactions and their corresponding enthalpy changes.

First, we'll need to multiply the second reaction by a factor of 2 to match the number of moles of O2 in the desired reaction:

2(SO2(g) + ½ O2(g) → SO3(g)) ΔH = 2(-141 kJ/mol)

Next, we'll need to multiply the first reaction by a factor of 2 to match the number of moles of SO2 in the second reaction:

2(S(s) + O2(g) → SO2(g)) ΔH = 2(-297 kJ/mol)

Now, let's rearrange the two equations to match the desired reaction:

2(SO2(g) + ½ O2(g)) ΔH = 2(-141 kJ/mol)
2SO2(g) + O2(g) → 2SO3(g)

2(S(s) + O2(g)) ΔH = 2(-297 kJ/mol)
2S(s) + 2O2(g) → 2SO2(g)

Finally, add the two equations together to cancel out the intermediates and obtain the desired reaction:

2SO2(g) + O2(g) + 2S(s) + 2O2(g) → 2SO3(g) + 2SO2(g)
2S(s) + 3O2(g) → 2SO3(g) + 2SO2(g)

Now, we can sum up the enthalpy changes:

2(-141 kJ/mol) + 2(-297 kJ/mol) = -282 kJ/mol + (-594 kJ/mol) = -876 kJ/mol

Therefore, the enthalpy change for the reaction
S(s) + 3/2O2(g) → SO3(g) is -876 kJ/mol.

To calculate the ΔH for the reaction: S(s) + 3/2O2(g) → SO3(g), we can use the Hess's Law of heat summation. According to Hess's Law, if a reaction can be expressed as the sum of two or more reactions, the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual reactions.

In this case, we can see that the desired reaction can be obtained by adding the first two reactions together, with appropriate coefficients to cancel out the common species.

1. Multiply the second reaction by 2 and reverse it to match the reaction form: 2(SO2(g) + ½O2(g) → SO3(g)), so that the O2(g) and SO3(g) cancel out in the overall reaction.

2. Now, add the two reactions together, including their enthalpy changes. The leftover species, S(s) and O2(g), will be the reactants in the overall reaction, and SO3(g) will be the product.

Adding the reactions:
S(s) + O2(g) → SO2(g) ΔH = -297 kJ/mol
2(SO2(g) + ½O2(g) → SO3(g)) ΔH = -2 * (-141 kJ/mol)

The O2(g) term cancels out, leaving us with the overall reaction:
S(s) + 3/2O2(g) → SO3(g)

Now, we can calculate the ΔH for the overall reaction by summing the enthalpy changes:
ΔH_overall = ΔH1 + ΔH2

ΔH_overall = (-297 kJ/mol) + (-2 * (-141 kJ/mol))

Calculating the values:
= -297 kJ/mol + 282 kJ/mol
= -15 kJ/mol

Therefore, the ΔH for the reaction S(s) + 3/2O2(g) → SO3(g) is -15 kJ/mol.