given `H_(2)O(l) to H^(+)(aq) + OH^(-)(aq),DeltaH= 57.32 kJ` `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l),DeltaH =-286.02 kJ` then , calculate the enthalpy of

Question

given `H_(2)O(l) to H^(+)(aq) + OH^(-)(aq),DeltaH= 57.32 kJ` `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l),DeltaH =-286.02 kJ` then , calculate the enthalpy of

given
`H_(2)O(l) to H^(+)(aq) + OH^(-)(aq),DeltaH= 57.32 kJ`
`H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l),DeltaH =-286.02 kJ`
then , calculate the enthalpy of formation of `OH^(-) at 25^(@)C`
A. `-228.8kJ`
B. `-343.52 kJ`
C. `+228.8 kJ`
D. `+ 343. 52 kJ`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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

Correct Answer - a
consider the formation of `H_(2)O`
`H_(2)(g)+1/2O_(2)(g) to H_(2)O(l),DeltaH=-286.20 kJ`
`DeltaH_(r)= DeltaH_(t)[H_(2)O(l)]-DeltaH_(t)[H_(2)(g))]-1/2 DeltaH_(t)[O_(2)(g)] `
` DeltaH_(t)(H_(2)O(l))=-286.20`
now consdier the ionisation of `H_(2)O`
`H_(2)O(l) to H^(+)(aq)+ OH^(-)(aq), DeltaH = 57.32 KJ H_(2)O(l) to H^(+)(aq)+ OH^(-)(aq), DeltaH = 57.32 KJ`