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Calculate the enthalpy change when 50 mL...

Calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)_(2)` reacts with `25mL` of `0.01 M HCI`. Given that `DeltaH^(Theta)` neutralisaiton of strong acid and string base is `140 kcal mol^(-1)`

A

`14 cal`

B

`35 cal`

C

`10 cal`

D

`7.5 cal`

Text Solution

Verified by Experts

The correct Answer is:
2
`m.eq.` of base `=50xx0.01xx2=1`
`m.eq.` of acid `=25xx0.01" "0.25`
`:. Me.eq.` of base reacted `=0.25`
`:.` energy released `=(140)/(1000)xx0.25Kcal=(140xx0.25xx1000)/(1000)cal=35cal`.
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