Chemistry - Thermochemistry Question with Solution | TestHub

$\text{2C}\left(\text{graphite}\right)+\text{3}{\text{H}}_2\left(\mathrm{g}\right)\to \text{1}{\text{C}}_2{\mathrm{H}}_6+\mathrm{\Delta }{\mathrm{H}}_{\mathrm{f}}\left({\mathrm{C}}_2{\mathrm{H}}_6\right)$
(A)${\mathrm{C}}_2{\mathrm{H}}_6+\frac{7}{2}{\mathrm{O}}_2\to \text{2C}{\text{O}}_2+\text{3}{\text{H}}_2\mathrm{O}$
$\mathrm{\Delta }{\mathrm{H}}_{\mathrm{C}}=–\text{1560}$
(B)${\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}$
$\mathrm{\Delta }{\mathrm{H}}_{\mathrm{C}}=–\text{393}.5$
(C) C(graphite)$+{\mathrm{O}}_2\to \text{C}{\text{O}}_2$
$\mathrm{\Delta }{\mathrm{H}}_{\mathrm{C}}=–\text{286kJ/mol}$
$\mathrm{\Delta }{\mathrm{H}}_{\mathrm{f}}\left({\mathrm{C}}_2{\mathrm{H}}_6\right)=3(–\text{393}.5)+2(–\text{286})+\text{156}0$
$=–\text{118}0.5+(–\text{572})+\text{156}0$
$=\text{192}.5$
$2\mathrm{C}\left(\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}\mathrm{i}\mathrm{t}\mathrm{e}\right)+3{\mathrm{H}}_2\to {\mathrm{C}}_2{\mathrm{H}}_6$(1) eqⁿ
${\mathrm{C}}_2{\mathrm{H}}_6+\frac{7}{2}{\mathrm{O}}_2\to 2\mathrm{C}{\mathrm{O}}_2+3{\mathrm{H}}_2\mathrm{O}$(2) eqⁿ
${\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2\to 2\mathrm{C}{\mathrm{O}}_2+3{\mathrm{H}}_2\mathrm{O}$(3) eqⁿ
$\mathrm{C}\left(\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}\mathrm{i}\mathrm{t}\mathrm{e}\right)+{\mathrm{H}}_2\mathrm{O}\to \mathrm{C}{\mathrm{O}}_2$(4) eqⁿ
To get equation (1) eqⁿ(2) + 2 × eqⁿ(3) 2 × eqⁿ
$∆{\mathrm{H}}_{\mathrm{f}}$of ethane$=1560+2\times \left(-286\right)+3(-393.5)$
$=1560-572-1180.5=192.5$