Question

For the reaction, $1\, g$ mole of $CaCO _{3}$ is enclosed in $5 \,L$ container
$CaCO _{3}(s) \longrightarrow CaO (s)+ CO _{2}(g)$
$K_{p}=1.16$ at $1073\, K$ then per cent dissociation of $CaCO _{3}$ is

• A. zero
• B. 6.58%
• C. 65%
• D. 100%

Answer 1

Answer: (B)

$CaCO _{3}( s ) \longrightarrow CaO ( s )+ CO _{2}(g)$
$k_{s}= P_{CO_3}$ (only gaseous molecule count)
$K_{p}=K_{c}(R T)^{\Delta n} $
$1.16 =\frac{x}{5}(0.0821 \times 1073)^{1}$
$ x =\frac{1.16 \times 5}{0.0821 \times 1073}=0.0658$
$=0.0658 \times 100=6.58 \%$