Question

I'm not 100% sure of the answer. Explanations are highly appreciated.

Answer 1

The correct answer is

B.

`\Delta H=2(1072)+498 -4(799)`

Step-by-step explanation:

Enthalpy of reaction :

It is the amount of energy released/absorbed when one mole of the substance is formed from the reactant at a constant pressure.

The enthalpy of a reaction can be calculated using :

`\Delta H=\Delta H_(reactants)-\Delta H_(products)`

`2CO+O_(2)\rightarrow 2CO_(2)`

`\Delta H_(reactants)=2(C\equiv O)+O=O`

`\Delta H_(reactants)=2(1072)+498`

`H_(products)=2(2* C=O)`

`H_(products)=4* 799`

`\Delta H=\Delta H_(reactants)-\Delta H_(products)`

`\Delta H=2(1072)+498 -4(799)`

`units = (kJ)/(mole)`

Please note that :

The carbon monoxide , CO should be taken as C triple bond O. Not C=O .

So , the bond energy =1072 is used

`\Delta H=2(1072)+498 -4(799)`