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Using the Gibbs energy change Delta G^(...

Using the Gibbs energy change `Delta G^(@) = +63. 3 kJ ` for the following reaction
` Ag_2CO_3( s) hArr 2Ag^(+) (aq) +CO_3^(2-) (aq) "the " K_(sp) " of " Ag_2CO_3(s)`
in water at ` 25^(@) C ` is (R = 8.314 `JK^(-1) " mol ^(-1))`

A

`3.2xx10^(-26)`

B

`8.0xx10^(-12)`

C

`2.9xx10^(-3)`

D

`7.9xx10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaG^@` is related to `K_sp` by the equation. `DeltaG^@=-2.303"RT log "K_sp`
Given, `Deltag^@=+63.3kJ`
`=63.3xx10^3J`
Thus, substance `DletaG^@=63.3xx10^3J`.
`R=8.314JK^(-1)mol^(-1)` and `T=298K[25+273K]` from the above equation we get, `63.3xx10^3=--2.303xx8.314xx298logK_sp`
`therefore" "log K_sp=-11.09`
`rArrK_sp ="anti log "(-11.09)`
`K_sp=8.0xx10^(-12)`
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