Calculate $\mathrm{\Delta H}$ in Joules for

$\mathrm{C}$ (graphite) $\to \mathrm{C}$ (Diamond)

by using the following data

$$\begin{gathered} \mathrm{C}\left(\text{ graphiter }\right)+{\mathrm{O}}_2( \mathrm{g})\to {\mathrm{CO}}_2( \mathrm{g}) \\ {\mathrm{\Delta H}}^0=-393.5\mathrm{KJ} \end{gathered}$$

$\mathrm{C}$ (Diamond) $+{\mathrm{O}}_2\left(\mathrm{ }\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{ }\mathrm{g}\right)$

${\mathrm{\Delta H}}^0=-395.4\mathrm{KJ}$

Given data –

We are given reactions and are asked to Calculate $\Delta \mathrm{H}$ in Joules based on these reactions.

Concept used –enthalpy of reaction

Detailed solution -

$$\begin{gathered} \mathrm{C}\left(\mathrm{s}\text{, graphite }\right)+{\mathrm{O}}_2( \mathrm{g})\to {\mathrm{CO}}_2( \mathrm{g})\Delta \mathrm{H}°=-393.5 \mathrm{k}\mathrm{J} \\ \mathrm{C}\left(\mathrm{s}\text{, diamond }\right)+{\mathrm{O}}_2( \mathrm{g})\to {\mathrm{CO}}_2( \mathrm{g})\Delta \mathrm{H}°=-395.4 \mathrm{k}\mathrm{J} \end{gathered}$$

Look at the reason is that the first, unreversed equation is exothermic. We know this $f$

$$\begin{gathered} {\mathrm{CO}}_2( \mathrm{g})+\mathrm{C}\left(\mathrm{s}\text{, graphite }\right)+{\mathrm{O}}_2( \mathrm{g})\to {\mathrm{CO}}_2( \mathrm{g})+\mathrm{C}\left(\mathrm{s}\text{, diamond }\right)+{\mathrm{O}}_2( \mathrm{g}) \\ \Delta \mathrm{H}°=(-393.5 \mathrm{k}\mathrm{J})+(+395.4 \mathrm{k}\mathrm{J})=1.9kJ=1900J \end{gathered}$$

Hence option A  is the correct answer.