Chemical Equilibrium — Chemistry STD 11 Science — Question
Maharashtra BoardEnglish MediumSTD 11 ScienceChemistryChemical Equilibrium6 Marks
Question
The unit of $KC$ is different for different reactions. Explain this statement with suitable examples.
✓
Answer
Unit of equilibrium constant:
$i$. The unit of equilibrium constant depends upon the expression of $K _{ C }$ which is different for different equilibria. Therefore, the unit of $K _{ C }$ is also different for different reactions.
$ii$. Consider the following equilibrium reaction: $H _{2( g )}+ I _{2( g )} \rightleftharpoons \ce2 HI _{( g )}$
$K _{ C }=\frac{\left[ HI _{( g )}\right]^2}{\left[ \ce{H _{2( g )}}\right)\left[ I _{2( g )}\right]}$Concentration is expressed in mol $dm ^{-3}$.
$\therefore \text { Units of } K _C =\frac{\left[ \text{moldm} ^{-3}\right]^2}{\left[ \text{moldm} ^{-3}\right]\left[ \text{moldm} ^{-3}\right]}$
$ =\frac{\left[ \text{moldm} ^{-3}\right]^2}{\left[ \text{moldm} ^{-3}\right]^2} \text { }$ As all the units cancel out, $K _C$ has no units.
$iii$. Consider the following equilibrium reaction: $\ce{N _{2(g)}+3 H _{2(g)}} \rightleftharpoons 2 \ce{NH _{3(g)}}$
$K _{ C }= \frac{\left[ NH _{3(g) }\right)^2}{\left[ N _{2(g) }\right)\left[ H _{2(g)}\right]^3}$
$\text { Unit of } K _{ C } =\frac{\left[ \text{moldm} ^{-3}\right]^2}{\left[ \text{moldm} ^{-3}\right]\left[ \text{moldm} ^{-3}\right]^3}$
$ =\frac{\left[ \text{moldm} ^{-3}\right]^2}{\left[ \text{moldm} ^{-3}\right]^4}$
$ =\left[ \text{moldm} ^{-3}\right]^{-2}$
$ = mol ^{-2} dm ^6 \text { }$
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