Question

In the gas phase reaction ${{C}_{2}}{{H}_{4}}+{{H}_{2}}\Leftrightarrow {{C}_{2}}{{H}_{6}}$ the equilibrium constant can be expressed in units of:
A. $Lt{{r}^{-1}}mol{{e}^{-1}}$
B. $Ltr.Mol{{e}^{-1}}$
C. $Mol{{e}^{2}}Ltr{{.}^{-2}}$
D. $Molelt{{r}^{-1}}$

Answer 1

The equilibrium constant expresses the relationship between the reactants and products with the powers raised in the form of moles with different units as it changes with the reaction. It is donated by K.

Complete step by step answer:
${{C}_{2}}{{H}_{4}}+{{H}_{2}}\Leftrightarrow {{C}_{2}}{{H}_{6}}$
The equilibrium constant K of this reaction will be written as:
$K=\dfrac{\left[ {{C}_{2}}{{H}_{6}} \right]}{\left[ {{C}_{2}}{{H}_{4}} \right]\times \left[ {{H}_{2}} \right]}$
These are all gases present here, so precisely the equilibrium constant will be written in the ${{K}_{p}}$ form. Now the units of equilibrium constant are expressed as ${{\left( moles/lit \right)}^{\Delta n}}$
Where, $\Delta n$ refers to the difference in moles of products and reactants, in this reaction the $\Delta n$ will be:
$\Delta n=1-\left( 1+1 \right) = -1$
Which means that ${{\left( moles/lit \right)}^{-1}}$.
Hence the answer of this question is B. $Ltr.Mol{{e}^{-1}}$.

Note: Please note that the answer of this question is B and not D because if you look at all the options, option A and C are eliminated at first glance because the mole of the reactants and product are only one, and in Option A, it means that both the litre and moles are in denominator which is not true that’s why it is also not correct. Now, if you look at option D.$Molelt{{r}^{-1}}$, it means that power of the (moles/litre) was 1 and when litre became the numerator it became -1 which is also wrong as the power of (moles/litre) is -1, when they are in multiplied form, we can reverse the moles, therefore, it becomes the $Ltr.Mol{{e}^{-1}}$.