JEE MAIN - Chemistry (2025 - 29th January Evening Shift - No. 15)
If $\quad C$ (diamond $) \rightarrow C$ (graphite) $+X \mathrm{~kJ} \mathrm{~mol}^{-1}$
C (diamond) $+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{Y} \mathrm{kJ} \mathrm{mol}{ }^{-1}$
C (graphite) $+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{Z} \mathrm{kJ} \mathrm{mol}^{-1}$
at constant temperature. Then
Explanation
Given,
$${C_{(diamond)}}\buildrel {} \over \longrightarrow {C_{(graphite)}} + X\,kJ\,mo{l^{ - 1}}$$ ............ (1)
$${C_{(diamond)}} + {O_2}(g)\buildrel {} \over \longrightarrow C{O_2}(g) + Y\,kJ\,mo{l^{ - 1}}$$ ............. (2)
$${C_{(graphite)}} + {O_2}(g)\buildrel {} \over \longrightarrow C{O_2}(g) + Z\,kJ\,mo{l^{ - 1}}$$ .............. (3)
Condition for temperature : constant
Hess's law is applied here. (Hess's law of constant heat summation)
The given reactions (2) and (3),
(2) - (3) $\Rightarrow$
$${C_{(diamond)}} + {O_2}(g) - \left( {{C_{(graphite)}} + {O_2}(g)} \right)\buildrel {} \over \longrightarrow C{O_2}(g) + Y - \left( {C{O_2}(g) + Z} \right)$$
$${C_{(diamond)}} + {O_2}(g) - {C_{(grpahite)}} - {O_2}(g)\buildrel {} \over \longrightarrow C{O_2}(g) + Y - C{O_2}(g) - Z$$
$${C_{(diamond)}} - {C_{(graphite)}} \to Y - Z$$
$${C_{(diamond)}} \to {C_{(graphite)}} + (Y - Z)$$ ............. (4)
Comparing (1) and (4),
$$X = Y - Z$$
So, the correct answer is option (2) $$X = Y - Z$$
Comments (0)
