MCQ
Combustion of $1$ mole of benzene is expressed at

$\mathrm{C}_6 \mathrm{H}_6(1)+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(1) \text {. }$

The standard enthalpy of combustion of $2 \mathrm{~mol}$ of benzene is - ' $x$ ' $k J$.

$\mathrm{x}=$. . . . . . . . . .

$(1)$ standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{C}_6 \mathrm{H}_6(1)$, for the reaction $6 \mathrm{C}$ (graphite) $+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_n(1)$ is $48.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

$(2)$ Standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{CO}_2(\mathrm{~g})$, for the reaction $\mathrm{C}$ (graphite) $+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_2(\mathrm{~g})$ is $-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

$(3)$ Standard and Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}(1)$, for the reaction $\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(1)$ is $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

  • $6535$
  • B
    $6540$
  • C
    $6545$
  • D
    $6550$

Answer

Correct option: A.
$6535$
a
$6 \mathrm{C} \text { (graphite) }+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\ell) ; \Delta \mathrm{H}=48.5 \mathrm{~kJ} / \mathrm{mol}$

$\mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta \mathrm{H}=-393.5 \mathrm{~kJ} / \mathrm{mol}$

$\mathrm{H}_2^{(\mathrm{g})}+\frac{1}{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286 \mathrm{~kJ} / \mathrm{mol}$

$\text { equation }-(1) \times 1+(2) \times 6+(3) \times 3$

$-48.5-6 \times 393.5-3 \times 286$

$=-3267.5 \mathrm{~kJ} \text { for } 1 \mathrm{~mol}$

$=-6535 \mathrm{~kJ} \text { for } 2 \mathrm{~mol}$

Ans. $6535 \mathrm{~kJ}$

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