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Assume the enthalpy of combustion of methane and $CH_3OH$ are x and y respectively. If the enthalpy of $CH_4 + \frac{1}{2}O_2 \rightarrow CH_3OH$ is negative, then which relation is correct?

(a) $x > y$ (b) $x < y$ (c) $x = y$ (d) $x \geq y$
Can you answer this question?
 
 

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Answer: $x > y$
 
On combustion of methane,
$CH_4 + O_2\rightarrow CO_2 + 2H_2O$ ; $\Delta H_1 = -x$ kJ ...(i)
Reaction for combustion of ethanal
$CH_3OH + \frac{1}{2}O_2 \rightarrow CO_2 + 2H_2O$; $\Delta H_2 = -y$ kJ ...(ii)
By subtracting (ii) from (i), we obtain,
$CH_4 + \frac{1}{2} O_2 \rightarrow CH_3OH$; $\Delta H_3 = -ve$
or $-x-(-y) = -ve$
or $y-x = -ve$
Hence, $x>y$
answered Feb 25, 2014 by mosymeow_1
 

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