Question

For the allotropic change represented by the equation:

$C\left(graphite\right)\to C\left(diamond\right),△H=1.9kJ.$.

If $6$g of diamond and $6$g of graphite are separately burnt to yield $C{O}_2$, the enthalpy liberated in first case is?

• A. less than in the second case by $1.9$ kJ
• B. more than in the second case by $11.4$ kJ
• C. more than in the second case by $0.95$ kJ
• D. less than in the second case by $11.4$ kJ

Answer 1

The correct option is C more than in the second case by $0.95$ kJ

Solution:- (C) more than in the second case by $0.95kJ$

No. of moles in $6g$ of graphite and diamond each $=0.5\text{mole}$

As we know that combustion is an exothermic reaction, thus the enthalpy of the combustion will be taken as negative.

Therefore,

Combustion of $0.5$ mole of graphite:-

$\frac{1}{2}{C}_{\text{graphite}}+\frac{1}{2}{O_2}_{\left(g\right)}⟶\frac{1}{2}{C{O}_2}_{\left(g\right)}\Delta H=-xkJ.....\left(1\right)$

Given:-

$C_{\text{graphite}}⟶C_{\text{diamond}}\Delta H=1.9kJ$

$C_{\text{diamond}}⟶C_{\text{graphite}}\Delta H=-1.9kJ$

$\frac{1}{2}\times [C_{\text{diamond}}⟶C_{\text{graphite}};\Delta H=-1.9kJ]$

$\frac{1}{2}{C}_{\text{diamond}}⟶\frac{1}{2}{C}_{\text{graphite}}\Delta H=-0.95kJ.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\frac{1}{2}{C}_{diamond}+\frac{1}{2}{O_2}_{\left(g\right)}+\frac{1}{2}{C}_{\text{graphite}}⟶\frac{1}{2}{C}_{\text{graphite}}+\frac{1}{2}{C{O}_2}_{\left(g\right)}\Delta H=-(x+0.95)kJ$

$\frac{1}{2}{C}_{diamond}+\frac{1}{2}{O_2}_{\left(g\right)}⟶\frac{1}{2}{C{O}_2}_{\left(g\right)}\Delta H=-(x+0.95)kJ$

Here $-ve$ sign indicates that the reaction is exothermic.

Hence the enthalpy liberated in the first case is more than in the second case by $0.95kJ$.