The entropies of `H_(2)(g) ` and `H(g) ` are 130.6 and `114.6J mol^(-1) K^(-1)` respectively at 298 K. Using the data given below calculate `DeltaH^(@

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The entropies of `H_(2)(g) ` and `H(g) ` are 130.6 and `114.6J mol^(-1) K^(-1)` respectively at 298 K. Using the data given below calculate `DeltaH^(@

The entropies of `H_(2)(g) ` and `H(g) ` are 130.6 and `114.6J mol^(-1) K^(-1)` respectively at 298 K. Using the data given below calculate `DeltaH^(@)` ( in kJ/ mol) of the reaction given below.
`H_(2)(g) to 2H(g)` , `DeltaG^(@) = 406.62 kJ//mol`

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 436KJ/mol
`DeltaS = (2 xx 114.6 )- 130.6=98.6 J`
`DeltaH = DeltaG + T DeltaS`
`= 436 kJ//mol`

4 views Answered 2023-02-05 15:29