MCQ
The heat of combustion of $C_xH_y$, carbon and hydrogen are $a, b$ and $c\  cal/mole$ respectively. The heat of formation of $C_xH_y$ will be 
  • A
    $-\left( {xb\, + \,\frac{{yc}}{2}\, + a} \right)cal$
  • $\left( {xb\, + \,\frac{{yc}}{2}\, - a} \right)cal$
  • C
    $\left( {xb\, - \,\frac{{yc}}{2}\, + a} \right)cal$
  • D
    $\left( {xb\, - \,\frac{{yc}}{2}\, - a} \right)cal$

Answer

Correct option: B.
$\left( {xb\, + \,\frac{{yc}}{2}\, - a} \right)cal$
b
We aim at the equation.

$xC + \frac{y}{2}{H_2} \to \mathop C\limits_x \,\mathop H\limits_y \,\,\,\,\,\,\,\,\Delta {H_f}$

We get, $\mathop {xC}\limits_{\Delta {H_C} = b}  + \mathop {\frac{y}{2}\,{H_2}}\limits_{\Delta {H_C} = c}  \to \mathop {\mathop C\limits_x \,\mathop H\limits_y }\limits_{\Delta {H_C} = a} $

Therefore, heat of formation

$\Delta H = \left( {xb + \frac{y}{2}c - a} \right)\,cal.$

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