Question

For the chemical equilibrium,
$CaC{O}_{3(s)}\leftrightharpoons Ca{O}_{(s)}+C{O}_{2(g)}$
$\Delta H_r^{\circ }$ can be determined from which one of the following plots?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

For the reaction,

$CaC{O}_{3(x)}\rightleftharpoons Ca{O}_{(s)}+C{O}_2(g)$

$K_p=p_{C{o}_2}$ and $K_{(:}=\left[C{O}_2\right]$

$(\because \left[CaC{O}_3\right]=1$ and $[CaO]=1$ for solids $]$

According to Arrhenius equation we have

$K=A{e}^{-\Delta H_r^{\circ }}/RT$

Taking logarithm, we have

$\log K_p=\log A-\frac{\Delta H_r^{\circ }}{RT(2.303)}$

This is an equation of straight line. When $\log K_p$, is plotted against $1/T$.

we get a straight line.

The intercept of this line $=\log A$, slope $=-\Delta H_r^{\circ }/2.303R$ Knowing the value of slope from the plot and universal gas constant $R,\Delta H_r^{\circ }$ can be calculated. (Equation of straight line : $Y=mx+C$. Here,

$\underset{Y}{\log K_p}=-\underset{m}{\frac{\Delta H_r^{\circ }}{2.303R}}\underset{x}{\left(\frac{1}{T}\right)}+\underset{C}{\mathrm{logA}}$