ThermochemistryHard

Question

Given that

C(s) + O2(g) → CO2(g); ΔHo = −X kJ

2CO(g) + O2(g) → 2CO2(g): ΔHo = −Y kJ

The enthalpy of formation of carbon monoxide will be

Options

A.(2X − Y)/2
B.(Y − 2X)/2
C.2X − Y
D.Y − 2X

Solution

$C(s) + O_{2}(g) \rightarrow CO_{2}(g);\Delta H^{o} = - x\text{ KJ}$

$- \frac{1}{2} \times \left\lbrack 2CO(g) + O_{2}(g) \rightarrow 2CO_{2}(g);\Delta H^{o} = - y\text{ kJ} \right\rbrack$

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$C(s) + \frac{1}{2}O_{2}(g) \rightarrow CO(g);\Delta H^{o} = ( - x) - \frac{1}{2}( - y) = \frac{y - 2x}{2}$

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