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(g))hArrCaO_((s)) + CO_(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 equition of straight line. When 1oh `K_(p)` is plotted against `1//t` . we get a straight line.

The intercept of this line `=log A`, slope `=- Delta H_(r)^(@)//2.303 R`
Knowing the value of slop from the plot and universal gas constant R, `Delta_(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