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K(p) for a reaction at 25^(@)C is 10 atm...

`K_(p)` for a reaction at `25^(@)C` is `10` atm. The activation energy for forward and reverse reactions are `12` and `20 kJ mol^(-1)` respectively. The `K_(c)` for the reaction at `40^(@)C` will be:

A

`4.33 xx 10^(-1)` M

B

`3.33 xx 10^(-2)` M

C

`3.33 xx 10^(-10)`M

D

`4.33 xx 10^(-2)` M

Text Solution

Verified by Experts

The correct Answer is:
C

Enthalpy changed of a reaction is given by
`DeltaH=E_(a(f))-E_(a(b))`
where `E_(a(f)) and E_(a(b))` are energies of activation for the forward and backward reactions.
`DeltaH=12 -20 =-8 kJ //mol`
`K_p` for the reaction at `25^@C -10` atm. Since `K_p` is expressed in atmoshere, `Deltan=+1`
`because K_p =K_c(RT)^(Deltan), K_c=(10)/(0.0821xx298)=0.4 M`
`K_c "at" 40^@C` is given by
`log""(k_c)_(40)/(K_c)_(25)=(DeltaH)/(2.303R)[1/(T_1)-1/(T_2)]=(-8xx1000)/(2.303xx8.314)xx(15)/(298xx313)=-0.06719`
`(K_c)_(40)//(K_c)_(25)=0.85`
`(K_c)_(40)=0.85xx0.4 =0.34M`
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