Question

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

Answer 1

Answer:

The correct answer is

B.

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

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 :

$\Delta H=\Delta H_{reactants}-\Delta H_{products}$

$2CO+O_{2}\rightarrow 2CO_{2}$

$\Delta H_{reactants}=2(C\equiv O)+O=O$

$\Delta H_{reactants}=2(1072)+498$

$H_{products}=2(2\times C=O)$

$H_{products}=4\times 799$

$\Delta H=\Delta H_{reactants}-\Delta H_{products}$

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

$units = \frac{kJ}{mole}$

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

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