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JEE Main 20207 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2020 · 7 January · Evening Shift · Chemistry · Thermodynamics (C)

The standard heat of formation ( ∆ _ f H _ 298 ^ 0 ) of ethane (in...

The standard heat of formation (∆fH2980)\left({∆}_{f}{H}_{298}^{0}\right) of ethane (in kJ/molkJ / mol ), if the heat of combustion of ethane, hydrogen and graphite are −1560,−393.5and−286kJ/mol- 1560, - 393.5 and - 286kJ / mol , respectively is ________

Correct answer

0

Step-by-step solution

2C(graphite)+3H2(g)→C2H6(g)2\text{C} ( \text{graphite} ) + 3{\text{H}}_{2} ( \text{g} ) \to {\text{C}}_{2} {\text{H}}_{6} ( \text{g} )

ΔfH(C2H6)=2ΔHcomb(Cgaphite)+3ΔHcomb(H2){\Delta}_{\text{f}} \text{H}\left({\text{C}}_{2} {\text{H}}_{6}\right) = 2\Delta{\text{H}}_{\text{comb}} \left({\text{C}}_{\text{gaphite}}\right) + 3\Delta{\text{H}}_{\text{comb}} \left({\text{H}}_{2}\right)

−ΔHcomb(C2H6)- \Delta{\text{H}}_{\text{comb}} \left({\text{C}}_{2} {\text{H}}_{6}\right)

=−(286×2)−(393.5×3)−(−1560)= - ( 286 \times 2 ) - ( 393.5 \times 3 ) - ( - 1560 )

=−572−1180.5+1560=−192.5kJ/mole= - 572 - 1180.5 + 1560 = - 192.5\text{kJ} / \text{mole}
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