Question

In the preparation of $CaO$ from $CaC{O}_3$ using the equilibrium
$CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$
$K_p$ is expressed as
$log{K}_p=7.282-\frac{8500}{T}$
During complete decomposition of $CaC{O}_3$ , the temperature in celdus to be used is

• A. 1167
• B. 894
• C. 8500
• D. 850

Answer 1

Answer: (B)

For the reaction,
$CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$
$K_p=p_{C{O}_2}$
$[\because$ Reaction takes place in open atmosphere,
$\therefore p_{C{O}_2}=1$]
or $K_p=1$
$\because log{K}_p=7.282-\frac{8500}{T}$
$\therefore log1=7.282-\frac{8500}{T}$
(where, $T=$ absolute temperature)
or $0=7.282-\frac{8500}{T}$
or $T=1167.261K$
$=(1167.26-273{)}^{\circ }C$
$=894.26^{\circ }C$
$\approx 894^{\circ }C$