Question

For the reaction ${{H}_{{{2}_{(g)}}}}+{{I}_{{{2}_{(g)}}}}2H{{I}_{(g)}}$at 721K the value of equilibrium constant ${{K}_{c}}$is 50. When the equilibrium concentration of both is 0.5M, value of \[{{K}_{p}}\] under the same conditions will be:
(A) 0.002
(B) 0.2
(C) 50.0
(D) $\dfrac{50}{RT}$

Answer 1

Hint: In order to solve this question, we will proceed by understanding that ${{K}_{c}}$is the equilibrium constant in terms of concentration whereas \[{{K}_{p}}\] is the equilibrium constant in terms of partial pressure. We’ll make use of the formula \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\]

Formula used: \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\] where \[{{K}_{p}}\] is the equilibrium constant in terms of partial pressure of reactant and product at equilibrium and ${{K}_{c}}$ is the equilibrium constant in terms of equilibrium concentration of product and reactant. R is the universal gas constant, T is the temperature and $\Delta {{n}_{g}}$is the difference of the number of moles of gases in product and reactant.

Complete Step by Step Solution:
According to the question the required chemical reaction at equilibrium is given as follows:
${{H}_{{{2}_{(g)}}}}+{{I}_{{{2}_{(g)}}}}2H{{I}_{(g)}}$
which shows that hydrogen gas and iodine gas are combining to form 2 moles of hydrogen iodide.
Using the formula:
\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\]

Given in the question:${{K}_{c}}=50$, T=721K and we know R is the universal gas constant.
To find the value of $\Delta {{n}_{g}}$ we will subtract the number of gaseous moles in reactant from the number of gaseous moles in the product.
$\Delta {{n}_{g}}=2$(no. of moles of $HI$) $-1$(no. of mole of ${{H}_{2}}$) $-1$(no. of moles of ${{I}_{2}}$)
Therefore, $\Delta {{n}_{g}}=2-1-1=0$

Putting the respective values in formula:
\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\]
${{K}_{p}}=50{{(RT)}^{0}}$
And we know anything raise to power zero is one
Therefore, \[{{K}_{p}}=50\]
Hence, the correct option is C. 50.0

Note: The most important thing to remember here is the relation between ${{K}_{p}}$ and ${{K}_{c}}$. The other thing to look for is the value of $\Delta {{n}_{g}}$, while calculating its value always consider the moles of gaseous molecules only on the both sides. Any liquid or solid molecule will not be considered for the calculation of the value of $\Delta {{n}_{g}}$.