Question

For the chemical equilibrium,
$CaCO _{3( s )} \leftrightharpoons CaO _{( s )}+ CO _{2( g )}$
$\Delta H_{r}^{\circ}$ can be determined from which one of the following plots?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

For the reaction,

$CaCO _{3(x)} \rightleftharpoons CaO _{(s)}+ CO _{2}(g)$

$K_{p}=p_{ Co _{2}}$ and $K_{(:}=\left[ CO _{2}\right]$

$\left(\because\left[ CaCO _{3}\right]=1\right.$ and $[ CaO ]=1$ for solids $]$

According to Arrhenius equation we have

$K=A e^{-\Delta H_{r}^{\circ}} /R T$

Taking logarithm, we have

$\log K_{p}=\log A-\frac{\Delta H_{r}^{\circ}}{R T(2.303)}$

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 $=-\Delta H_{r}^{\circ} / 2.303 R$ Knowing the value of slope from the plot and universal gas constant $R, \Delta H_{r}^{\circ}$ can be calculated. (Equation of straight line : $Y=m x+C$. Here,

$\underset{Y}{\log K_{p}}=-\underset{m}{\frac{\Delta H_{r}^{\circ}}{2.303 R}}\underset{x}{\left(\frac{1}{T}\right)}+\underset{C}{\log A}$