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Answer (Detailed Solution Below)

Option 2 : −726.5 kJ mol−1

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Detailed Solution

सही उत्तर −726.5 kJ mol−1 है।

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PrepSnapprepsnap.in • Part -A

Question:

Given $\Delta_f H^\circ$ values (kJ mol$^{-1}$): CO2(g) = −393.5, H2O(l) = −285.8 and CH3OH(l) = −238.6, the standard enthalpy of combustion of liquid methanol is / दिए गए $\Delta_f H^\circ$ मानों (kJ mol$^{-1}$) से: CO2(g) = −393.5, H2O(l) = −285.8 तथा CH3OH(l) = −238.6, द्रव मेथेनॉल की मानक दहन एन्थैल्पी है

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−1074.7 kJ mol−1
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−726.5 kJ mol−1
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+726.5 kJ mol−1
4.
−964.3 kJ mol−1

Correct Answer: −726.5 kJ mol−1

For CH3OH + 3/2 O2 → CO2 + 2H2O, ΔcH° = [ΔfH°(CO2) + 2ΔfH°(H2O)] − ΔfH°(CH3OH) = (−393.5 − 571.6) + 238.6 = −726.5 kJ mol−1. / CH3OH + 3/2 O2 → CO2 + 2H2O हेतु ΔcH° = [ΔfH°(CO2) + 2ΔfH°(H2O)] − ΔfH°(CH3OH) = (−393.5 − 571.6) + 238.6 = −726.5 kJ mol−1।

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