# Enthalpy change of reaction

Suppose you have the following equation:

$$\ce{2NH4SCN(s) + Ba(OH)2(s) ->Ba(SCN)2(s) + 2H2O(l) + 2NH3(g)}$$

You initiate this reaction using 15.22g of the $\ce{NH4SCN}$ (and excess of the barium hydroxide) in a boiling tube resting in 50g of water. The temperature rises by 11 degrees. Therefore energy = $4.2 \times 11 \times 50 = 2.31kJ$ and moles = $\frac{15.22}{76.1} = 0.2mol$

I would then assume that it would just be $\frac{2.31}{0.2} = 11.55kJmol^{-1}$

But the actual answer is 23.1 (double my answer) which I am guessing is because there are 2 moles of $\ce{2NH4SCN}$. I do not understand why this is the case, because you are using 0.2 moles and the energy raised is 2.31 so why the multiplication by 2?

• What is 15.22 g?
– LDC3
May 25, 2014 at 20:44
• sorry i have updated the question May 25, 2014 at 20:44

## 1 Answer

The reason lies with the definition of the Heat of Reaction. It states that it is the amount of heat when 1 mole of the product is formed. As you noted, since 2 moles of $\ce{NH4SCN}$ is needed for each mole of product, then you need to multiply by 2.

Added:
If you change the number of moles of $\ce{NH4SCN}$ to 0.1 moles (to remove the stoichiometric 2), then your calculations would give you the answer.

• I'm not sure I understand then what the moles value I get when I calculate mass divided by rel. molecular mass? - surely that would be the amount of moles used in the reaction? May 25, 2014 at 20:59
• never mind I understand! May 25, 2014 at 21:02
• I know how it works, but I have a hard time to express it. What you want is to have all the stoichiometry as 1. In order to keep a balanced equation, you need to modify the number of moles in the reaction (in this case you want half the ammonium cyanide, but the barium cyanide would be the same).
– LDC3
May 25, 2014 at 21:02