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Answer (Detailed Solution Below)

Option 3 : −890.3 kJ/mol

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Detailed Solution

सही उत्तर −890.3 kJ/mol है।

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PrepSnapprepsnap.in • Part -A

Question:

Given ΔHf° (kJ/mol): CO2(g) = $−393.5$, H2O(l) = $−285.8$, CH4(g) = $−74.8$. The enthalpy of combustion of methane is: / दिया है ΔHf° (kJ/mol): CO2(g) = $−393.5$, H2O(l) = $−285.8$, CH4(g) = $−74.8$। मेथेन की दहन एन्थैल्पी है:

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−965.1 kJ/mol
2.
−1040.0 kJ/mol
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−890.3 kJ/mol
4.
−815.5 kJ/mol

Correct Answer: −890.3 kJ/mol

For CH4 + 2O2 → CO2 + 2H2O, $\Delta H = [(-393.5) + 2(-285.8)] - (-74.8) = -965.1 + 74.8 = -890.3$ kJ/mol by Hess's law. / CH4 + 2O2 → CO2 + 2H2O के लिए हेस नियम से $\Delta H = [(-393.5) + 2(-285.8)] - (-74.8) = -965.1 + 74.8 = -890.3$ kJ/mol।

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