Q.

Heat of formation of H2Og at 1 atm and 250C is -243 kJ. ∆E for the reaction H2g+12 O2g   → H2Og  at 250C is

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a

241.8 KJ

b

-241.8KJ

c

-243 KJ

d

243 KJ

answer is B.

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Detailed Solution

∆H=∆E+∆nRT ∆n = -1/2 ∴ -243=∆E+1/2 × 8.314 × 298 × 10-3 ∴ ∆E=-241.76 KJ
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