Question

For the change, $C_{diamond}\to C_{graphite};\Delta H=-1.89kJ,$ if $6g$ of diamond and $6g$ of graphite are separately burnt to yield $C{O}_2$ , the heat liberated in first case is

• A. Less than in the second case by $1.89kJ$
• B. Less than in the second case by $11.34kJ$
• C. Less than in the second case by $14.34kJ$
• D. More than in the second case by $0.945kJ$

Answer 1

Answer: (D)

$C_{diamond}\to C_{graphite};\Delta H=-1.89kJ$ …(i)
Enthalpy of combustion of a substance is the change in enthalpy produced when one mole of the substance is completely burnt in air or oxygen at a given temperature.
$C_{diamond}+O_2\to C{O}_2;\Delta H=-q_1$ …(ii)
$C_{graphite}+O_2\to C{O}_2;\Delta H=-q_2$ …(iii)
By equations [(i) – (ii)]
$C_{diamond}\to C_{graphite};\Delta H=-q_1+q_2$
So $-q_1+q_2=-1.89$
or $q_2-q_1=-1.89$ for $12g$ $C_{diamond\to graphite}$
Thus, for $6g$ $C_{diamond\to graphite}$
$=\frac{-1.89}{2}=0.945kJ$ .