The calorific value of H2(g) at STP is 12.78 kJ/L hence approximate standard enthalpy of formation of H2O(l) is
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a
-143 kJ
b
-286 kJ
c
Zero
d
+286 kJ
answer is B.
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Detailed Solution
1 LH2(g) at STP=122.4mol.'. Heat released due to combustion of 122.4mol of H2(g) = 12.78 KJHeat released due to combustion of I mol of H2(g) = 12.78 × 22.4 = 286.27 kJ∴approximate standard enthalpy of formation of H2O(l) = 286 KJ