For the chemical equilibrium,
`CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)`
`Delta_(r)H^(ɵ)` can be determined from which one of the following plots?

Options

A.
B.
C.
D.

Answer

A

Explanation

For the reaction,
`CaCO_(3(s))hArrCaO(s) + C O_(2(g))`
`K_(p) = P_(CO_(2))` and `K_(C) = [CO_(2)]`
`.: [CaCO_(3)] = 1` and `[CaO] = 1` for solids]
According to Arrhenius equation we have
`K = Ae^(-Delta H_(r)^(@)//RT)`
Taking logarithm, we have
log `K_(p)` = log `A - (Delta H_(r)^(@))/(RT(2.303))`
This is an equilibrium of straight line. When log `K_(p)` is plotted against `1//T` , we get a straingth line.

The intercept of this line = log A, slope = `-Delta H_(r)^(@)//2.303R` Knowing the value of slope form the plot and universal gas constant R, `Delta H_(r)^(@)` can be calculated. Equation of straight line: `Y = mx + C` . Here,
log `K_(p) = -(Delta H_(r)^(@))/(2.303R) ((1)/(T))+ log A`
Y m x C