The heat of transition $(\Delta H_t)$ of graphite into diamond would be,where
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = x \ kJ \ mol^{-1}$
$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = y \ kJ \ mol^{-1}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = x \ kJ \ mol^{-1}$
$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = y \ kJ \ mol^{-1}$
- A$(x + y) \ kJ \ mol^{-1}$
- B$(x - y) \ kJ \ mol^{-1}$
- C$(y - x) \ kJ \ mol^{-1}$
- DNone of these
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Similar Questions
Given
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta_{r}H^{\theta} = -92.4 \, kJ \, mol^{-1}$
What is the standard enthalpy of formation of $NH_{3}$ gas?
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta_{r}H^{\theta} = -92.4 \, kJ \, mol^{-1}$
What is the standard enthalpy of formation of $NH_{3}$ gas?
Easy
View Solution$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)}$
$B.E. (H-H) = x_1$; $B.E. (O=O) = x_2$;
$B.E. (O-H) = x_3$
Heat of vaporisation of water $= x_4$,then $\Delta H_f$ [heat of formation of liquid water] is:
$B.E. (H-H) = x_1$; $B.E. (O=O) = x_2$;
$B.E. (O-H) = x_3$
Heat of vaporisation of water $= x_4$,then $\Delta H_f$ [heat of formation of liquid water] is:
Difficult
View SolutionIf $8.84 \ kJ$ heat is liberated for the formation of $3 \ g$ of ethane,calculate its $\Delta_{f} H^{\circ}$.
The enthalpy change for a reaction does not depend on:
Easy
View SolutionWhat will be the heat of formation of methane,if the heat of combustion of carbon is $-x \ kJ$,heat of formation of water is $-y \ kJ$ and heat of combustion of methane is $z \ kJ$?
DifficultAIIMS 2008
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