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The reaction CaCO3 hArr CaO(s) + CO2(g...

The reaction
`CaCO_3 hArr CaO(s) + CO_2(g)`
is in equilibrium in a closed vessel at 298 K . The partial pressure (in atm) of `CO_2` (g) in the reaction vessel is closest to :
[Given : the change in Gibbs energies of formation at 298 K and 1 bar for
`CaO(s) = - 603. 501 kJ mol^(-1)`
`CO_2(g) = -394. 389 kJ mol^(-1)`
`CaCO_(3)(s) = -1128. 79 kJ mol^(-1)`
Gas constant `R = 8.314 J K^(-1) mol^(-1)]`

A

`1.13 xx 10^(-23)`

B

`0.95`

C

`1.05`

D

`8.79 xx 10^(23)`

Text Solution

Verified by Experts

The correct Answer is:
A
`CaCO_3 (s) hArr CaO +CO_2 (g)`
`Delta_r G^@ = Delta G^@ (CaO ) + Delta_f G^@ (CO_2) - Delta _f G^@ (CaCO_3)`
`=- 603 .501 - 394 .389 + 1128 .79 = 130 .9 KJ mol^(-1)`
` Delta_r G^@ =- 2.303 RT log K_p`
` log K_P = ( 130.9 xx 1000)/(-2.303 xx 298 xx 8.314 ) =-22.94`
`K_p`= antilog` (-22.94 ) = 1.13 xx 10^(-23)`
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