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Variation of equiibrium constant K with ...

Variation of equiibrium constant `K` with temperature `T` is given by van `'t` Holf equation,
`log K=logA-(DeltaH^(@))/(2.303RT)`
A graph between log `K` and `T^(-1)` was a straight line as shown in the figure and having `theta=tan^(-1)(0.5)"and"OP=10`. Calculate:
(a) `DeltaH^(@)` (standard heat of reaction)when `T=300K,`
(b) `A` (pre-exponetial factor),
(c) Equilibrium constant `K` at `300 K,`
(d) `K "at" 900 K "if" DeltaH^(@)` is independent of temperature.

Text Solution

Verified by Experts

The correct Answer is:
(a) `9.574J mol^(-1);` (b)`A=10^(10);` (c) `9.96xx10^(9);` (d)`9.98xx10^(9)`
(a) `log_(10)K=log_(10)A-(DeltaH_(@))/(2.303RT)`
It is an equation of a straight line of the type `y=c+mx`
Slope 'm'=`tan0=(DeltaH^(@))/(2.303R)`
`0.5=(DeltaH^(@))/2.303xx8.314`
`DeltaH^(@)=9.574J mol^(-1)`
(b) `"Intercept",C'=log_(10)A=10` `A=10^(10)`
(c) `log K=10-(9.574)/(2.303xx8.314xx298)`
`K=9.96xx10^(9)`
(d) `log((K_(2))/(K_(1)))=(DeltaH)/2.303R {(1)/(T_(1))-(1)/(T_(2))}`
`log (K_(2))/(9.96xx10_(9))=(9.574)/(2.303xx8.314) {(1)/(298)-(1)/(798)}`
On solving `K_(2)=9.98xx10_(9)`
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