1.

On the basis of the following thermochemical data [(Delta_(f)G^(@)H^(+) (aq) = 0] H_(2)O (l) rarr H^(+) (aq), Delta H = 57.32 kJ H_(2) (g) + (1)/(2)O_(2) (g) rarr H_(2)O (l), Delta H = - 286 kJ The value of enthalpy of formation of OH^(-) at 25^(@)C is

Answer»

`- 22.88 kJ`
`- 228.88 kJ`
`228.88 kJ`
`- 343.52 kJ`

SOLUTION :`H_(2) (G) + (1)/(2)O_(2)(g) rarr H^(+) (aq) + OH^(-) (aq), Delta H_(1) = ?`
`H_(2)O (L) rarr H^(+) (aq) + OH^(-) (aq), Delta H_(2) = 57.32 kJ` …..(i)
`H_(2) (g) + (1)/(2)O_(2)(g) rarr H_(2)O (l), Delta H_(3) = - 286.2 kJ` …. (ii)
`Delta H_(1)` will be obtained by adding (i) and (ii)
`= 57.32 + (- 286.2) = - 228.88 kJ`


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