2010 Q16So they tell you the standard enthalpy of combustion of hydrogen is -286 kJmol^-1
If you write out the standard enthalpy of combustion of hydrogen you get:
H2(g) + 1/2 O2 (g) --> H2O (l)
Now if you look at the standard enthalpy of formation, you get the exact same equation (enthalpy change when 1 mole of a compound is formed from its elements, in their normal states, under standard conditions).
So the enthalpy change is exactly the same, -286 kJmol^-1, which is answer A.
2011 Q19The enthalpy change that they've asked you to find is from the top line to the bottom line of the thermochemical cycle, so to work it out you need to know all the values in the cycle. Therefore we need to determine the enthalpy change for the line which represents:
H2(g) + 1/2 O2 (g) --> H2O (l)
This represents the standard enthalpy of combustion of hydrogen, because it is the combustion of one mole of it in excess oxygen under standard conditions (cos all the chemicals are in their standard states).
If you check in the data book, you'll see that the standard enthalpy of combustion of hydrogen is -286 kJmol^-1.
Now you just need to plug all the numbers into your calculator, going round the cycle (arrows pointing up are positive, arrows pointing down are negative).
So you end up with:
-286 + 432 + 248 – 916 – 46 = -568 which is answer A.
Hope that helped
And to join in with the general chat, I find Unit 1 very boring, quite like Unit 2 and hate Unit 3 (although I am gradually coming to terms with it, which is a good job seeing as the exam is so very soon!).