Questions & Answers
Question
Answers
Related Questions
At a certain temperature and a total pressure of ${{10}^{5}}Pa$, iodine vapours contain $40%$ by volume of iodine atoms. ${{K}_{P}}$ for the equilibrium ${{I}_{2}}(g)2I(g)$.
A.$0.67$
B.$1.5$
C.$2.67\times {{10}^{4}}$
D.$9.0\times {{10}^{4}}$
Answer
![Verified]()
Verified
Verified
Hint: Here our reaction is homogeneous equilibrium of gaseous system so we can use equilibrium constant in terms of partial pressure as given we have iodine $40%$ which we can use to find the partial pressure of iodine atom.
Complete answer:
As we can express equilibrium constant of the reaction in terms of molar concentration of the reactant and product which we denote it as ${{K}_{C}}$.But for reaction involving gases which is given in this question it is more convenient to express the equilibrium constant in terms of partial pressure which we are denoting it as ${{K}_{P}}$.Sometimes these both equilibrium are same but not always.
To find the${{K}_{P}}$for the equilibrium ${{I}_{2}}(g)2I(g)$
Let us compare it with the general reaction $aA + bB and cC + dD$
For which ${{K}_{P}}=\dfrac{(p_{C}^{c})(p_{D}^{d})}{(p_{A}^{a})(p_{B}^{b})}$
${{K}_{P}}$$=\dfrac{{{[C]}^{c}}{{[D]}^{d}}{{(RT)}^{(c+d)}}}{{{[A]}^{a}}{{[B]}^{b}}{{(RT)}^{(a+b)}}}$
So no according to given equilibrium we can say that${{K}_{P}}=\dfrac{{{(p_{I}^{{}})}^{2}}}{(p_{{{I}_{2}}}^{{}})}$
The partial pressure of iodine atom will be:
$({{P}_{I}})=\dfrac{{{10}^{5}}\times 40}{100}$
$({{P}_{I}})=0.4\times {{10}^{6}}Pa$
Now the partial pressure of ${{I}_{2}}$ molecule will be:
$({{P}_{{{I}_{2}}}})=\dfrac{{{10}^{5}}\times 60}{100}$
$({{P}_{{{I}_{2}}}})=0.6\times {{10}^{5}}Pa$
Now putting values in the equation i.e.
${{K}_{P}}=\dfrac{{{(p_{I}^{{}})}^{2}}}{(p_{{{I}_{2}}}^{{}})}$
${{K}_{P}}=\dfrac{{{(0.4\times {{10}^{6}}Pa)}^{2}}}{0.6\times {{10}^{5}}Pa}$
Here we get the answer as ,
${{K}_{P}}=2.67\times {{10}^{4}}$
It must be remembered that for the existence of heterogeneous equilibrium pure solid and liquids must also be present at equilibrium, but their concentration or partial pressure do not appear in the expression of the equilibrium constant.
So, the correct answer is “Option C”.
Note: It is important that while calculating the value of ${{K}_{P}}$ the pressure should be expressed in the bar because the standard state for pressure is 1 bar.
1$pascal$, $Pa=1N{{m}^{-2}}$and $1bar={{10}^{5}}Pa$.
Equilibrium constant can also be expressed as dimensionless quantities if the standard state of reactants and products are specific. Like for example a pure gas has a standard state as 1 bar.
Complete answer:
As we can express equilibrium constant of the reaction in terms of molar concentration of the reactant and product which we denote it as ${{K}_{C}}$.But for reaction involving gases which is given in this question it is more convenient to express the equilibrium constant in terms of partial pressure which we are denoting it as ${{K}_{P}}$.Sometimes these both equilibrium are same but not always.
To find the${{K}_{P}}$for the equilibrium ${{I}_{2}}(g)2I(g)$
Let us compare it with the general reaction $aA + bB and cC + dD$
For which ${{K}_{P}}=\dfrac{(p_{C}^{c})(p_{D}^{d})}{(p_{A}^{a})(p_{B}^{b})}$
${{K}_{P}}$$=\dfrac{{{[C]}^{c}}{{[D]}^{d}}{{(RT)}^{(c+d)}}}{{{[A]}^{a}}{{[B]}^{b}}{{(RT)}^{(a+b)}}}$
So no according to given equilibrium we can say that${{K}_{P}}=\dfrac{{{(p_{I}^{{}})}^{2}}}{(p_{{{I}_{2}}}^{{}})}$
The partial pressure of iodine atom will be:
$({{P}_{I}})=\dfrac{{{10}^{5}}\times 40}{100}$
$({{P}_{I}})=0.4\times {{10}^{6}}Pa$
Now the partial pressure of ${{I}_{2}}$ molecule will be:
$({{P}_{{{I}_{2}}}})=\dfrac{{{10}^{5}}\times 60}{100}$
$({{P}_{{{I}_{2}}}})=0.6\times {{10}^{5}}Pa$
Now putting values in the equation i.e.
${{K}_{P}}=\dfrac{{{(p_{I}^{{}})}^{2}}}{(p_{{{I}_{2}}}^{{}})}$
${{K}_{P}}=\dfrac{{{(0.4\times {{10}^{6}}Pa)}^{2}}}{0.6\times {{10}^{5}}Pa}$
Here we get the answer as ,
${{K}_{P}}=2.67\times {{10}^{4}}$
It must be remembered that for the existence of heterogeneous equilibrium pure solid and liquids must also be present at equilibrium, but their concentration or partial pressure do not appear in the expression of the equilibrium constant.
So, the correct answer is “Option C”.
Note: It is important that while calculating the value of ${{K}_{P}}$ the pressure should be expressed in the bar because the standard state for pressure is 1 bar.
1$pascal$, $Pa=1N{{m}^{-2}}$and $1bar={{10}^{5}}Pa$.
Equilibrium constant can also be expressed as dimensionless quantities if the standard state of reactants and products are specific. Like for example a pure gas has a standard state as 1 bar.
Like this
Liked
-
LIVECourses
-
V PRO
-
JEE
-
JEE Main & Advanced 2020
-
JEE Main & Advanced 2021
-
JEE Main & Advanced 2022
-
-
NEET
-
NEET 2020
-
NEET 2021
-
NEET 2022
-
-
CBSE
-
Class 6
-
Class 7
-
Class 8
-
Class 9
-
Class 10
-
Class 11
-
Class 12
-
-
ICSE
-
Class 9
-
Class 10
-
-
Maharashtra Board
-
Class 9
-
Class 10
-
-
Micro Courses
-
NDA
-
-
PREMIUM1-on-1 Classes
-
Study Material
-
NCERT Solutions
-
Previous Year Papers
-
Mock Tests
-
Sample Papers
-
Reference Book Solutions
-
ICSE Solutions
-
School Syllabus
-
Revision Notes
-
Important Questions
-
Math Formula Sheets
-
-
More
Book your FREE Online counselling session
Share your contact information
Your FREE Online counselling session has been booked!
Vedantu academic counsellor will be calling you shortly for your Online Counselling session.
DONE
Please try again later
OK
NCERT Book Solutions
Reference Book Solutions
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
(Mon-Sun: 9am - 11pm IST)
Questions?
Chat with us
