From given following equations and $\Delta \mathrm{H}^{\circ}$ values, determine the enthalpy of reaction at…
$\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+6 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CF}_{4}(\mathrm{~g})+4 \mathrm{HF}(\mathrm{g})$
$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HF}(\mathrm{g}) ; \Delta \mathrm{H}_{1}^{\circ}=-537 \mathrm{~kJ}$
$\mathrm{C}(\mathrm{s})+2 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CF}_{4}(\mathrm{~g}) ; \Delta \mathrm{H}_{2}^{\circ}=-680 \mathrm{~kJ}$
$2 \mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g}) ; \Delta \mathrm{H}_{3}^{\circ}=52 \mathrm{~kJ}$
- $-1165$
- $-2486$
- $+1165$
- $+2486$
Solution
Asked in: JEE-TOPICTESTS-CHEMISTRY