# Calculating Enthalpy Change

Calculate the Change in Hf for sulfur dioxide, SO2 from its elements sulfur and oxygen. Use the balanced chemical equation and the following information.

S(s) + O2(g) -> SO2(g)

S(s) + 3/2 O2(g) -> SO3 Change in Hf = -395.7 kj/mol

2SO2(g) + O2 -> 2SO3(g) change in Hf = -198.4 kj/mol

A detailed explanation would be super helpful cause for some reason I struggle with this concept and any shortcuts just throw me off >.

Thanks!

kingchemist

Sun, 2009-11-29 06:02

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## Re: Calculating Enthalpy Change

Write the 'unknown' equation

S(s) + O2(g) -> SO2(g) [delta]H = ?

Write the 'known' equations and number them

1. S(s) + 3/2 O2(g) -> SO3 Change in Hf = -395.7 kj/mol

2. 2SO2(g) + O2 -> 2SO3(g) change in Hf = -198.4 kj/mol

Now look at the 'known' equations and decide what you need from them to make the 'unknown' equation. You can double, treble, reverse the equation ( I call this 'manipulating' the equations). If you reverse the equation, change the sign of [delta]H. If you double the equation, double the [delta]H value etc

From 1. we want S and O2 and we need them on the LHS to make the 'unknown' equation. We need 1 mole of S so this is the correct quantity, so we don't do anything to the quantity. Don't worry about O2 as this is involved in both reactions and will deal with itself.

From 2. we want SO2 on the RHS in the 'unknown' equation. So we need to reverse 2. and also divide it by 2 to get 1 mole of SO2

We say - divide 2. by 2, reverse 2 and add it to 1. (I rewrite the equation according to the manipulation - I know it takes longer but I think it helps.

S(s) + 3/2 O2(g) -> SO3 [delta]H

_{f}= -395.7 kJ/molSO3(g) -> SO2(g) + 1/2O2 [delta]H = +99.2 kJ/mol (notice the change in sign and halving of the value)

Adding the equations, we add both LHS to make the new LHS and do the same with RHS

S(s) + 3/2 O2(g) + SO3(g) -> SO3 + SO2(g) + 1/2O2 [delta]H = (-395.7 + (+99.2) kJ

Simplifying (cancellling quantites that appear on both sides of the equation.)

S(s) + O2(g) -> SO2(g) [delta]H = (-395.7 + (+99.2) kJ This is the unknown equation.

Loraleena

Sun, 2009-11-29 10:05

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## Re: Calculating Enthalpy Change

Thank you tons!!! This makes perfect sense now!!!!