The heat of transition ($\Delta H_t$) of graphite into diamond would be, given the following thermochemical eq

Question

The heat of transition ($\Delta H_t$) of graphite into diamond would be, given the following thermochemical equations:
1. C(graphite) + O$_2$(g) $\rightarrow$ CO$_2$(g); $\Delta H = x$ kJ
2. C(diamond) + O$_2$(g) $\rightarrow$ CO$_2$(g); $\Delta H = y$ kJ

Accepted Answer

To determine the heat of transition for the reaction C(graphite) $\rightarrow$ C(diamond), we apply **Hess’s Law of Constant Heat Summation**. This law states that the total enthalpy change for a reaction is the same whether it occurs in one step or multiple steps .

We are given two combustion reactions:
(i) C(graphite) + O$_2$(g) $\rightarrow$ CO$_2$(g); $\Delta H_1 = x$ kJ 
(ii) C(diamond) + O$_2$(g) $\rightarrow$ CO$_2$(g); $\Delta H_2 = y$ kJ

To find the enthalpy change for the transition from graphite to diamond, we subtract the second equation from the first:
[C(graphite) + O$_2$(g)] - [C(diamond) + O$_2$(g)] $\rightarrow$ [CO$_2$(g)] - [CO$_2$(g)]
C(graphite) - C(diamond) $\rightarrow$ 0
C(graphite) $\rightarrow$ C(diamond)

The enthalpy change for this process ($\Delta H_t$) is calculated by performing the same operation on the enthalpy values:
$$\Delta H_t = \Delta H_1 - \Delta H_2 = x - y$$

Thus, the heat of transition is **$(x - y)$ kJ mol$^{-1}$**.

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