Question

2A + C + D + EDetermine the enthalpy(AH)in kJ of the above reaction using the reactions below:1B → C AH = 415.3 kJA + 1B →D AH = -95.7 kJ6A ->6E AH = 207.9 kJ

Answer 1

Answer: ∆H total= -476.35kJ.

Explanations:

• We will follow Hess's Law of Constant Heat Summation:

(that states that if a reaction occurs in more than one route, then the the total enthalpy change for the reaction is the sum of all changes.)

For reaction :

`\begin{gathered} B\Rightarrow C\text{ }\Delta H\text{ = 415.3 kJ }\ldots\text{.}\mathrm{}(\text{ reaction 1)} \\ A+B\text{ }\Rightarrow\text{ D }\Delta H\text{ = -95.7kJ}\ldots.(\text{ reaction 2 ) } \\ 6A\text{ }\Rightarrow6E\text{ }\Delta H\text{ = 207.9kJ }\ldots\ldots(\text{reaction 3 ) } \end{gathered}`

∆H total =( reaction 3 )/6 +reaction 2 - reaction 1

this can be expressed as :

`\begin{gathered} \Delta H_{total\text{ }}=\text{ }\frac{207.9}{6\text{ }}+(-95.7\text{ ) - 415}.3\text{ } \\ \text{ = 34.65 -95.7 -415.3} \\ \text{ =-476.35kJ} \end{gathered}`

This means that ∆H total= -476.35kJ.