Concepts and reason
Equilibrium constant is the ratio between the concentration of products to the concentration of reactants. The equilibrium constant is expressed asK {\rm{K}} K . If the value of K {\rm{K}} K is more than 1, the equilibrium favors products and if the value of K {\rm{K}} K is less than 1, the equilibrium favors reactants.
Fundamentals
For a simple equilibrium reaction,
The equilibrium constant with respect to molar concentration is represented asK c {\rm{Kc}} K c .
K c = [ C ] c [ D ] d [ A ] a [ B ] b {\rm{Kc}}\,\,{\rm{ = }}\,\,\frac{{{{\left[ {\rm{C}} \right]}^{\rm{c}}}{{\left[ {\rm{D}} \right]}^{\rm{d}}}}}{{{{\left[ {\rm{A}} \right]}^{\rm{a}}}{{\left[ {\rm{B}} \right]}^{\rm{b}}}}} K c = [ A ] a [ B ] b [ C ] c [ D ] d
The equilibrium constant in terms of partial pressure can be represented asK p {{\rm{K}}_{\rm{p}}} K p .
K P = ( C ) c ( D ) d ( A ) a ( B ) b {{\rm{K}}_{\rm{P}}}\,{\rm{ = }}\,\,\frac{{{{\left( {\rm{C}} \right)}^{\rm{c}}}{{\left( {\rm{D}} \right)}^{\rm{d}}}}}{{{{\left( {\rm{A}} \right)}^{\rm{a}}}{{\left( {\rm{B}} \right)}^{\rm{b}}}}} K P = ( A ) a ( B ) b ( C ) c ( D ) d
The relation between both K c {\rm{Kc}} K c and K p {{\rm{K}}_{\rm{p}}} K p is:
K p = K c ( R T ) Δ n K p = e q u i l i b r i u m c o n s t a n t i n t e r m s o f p a r t i a l p r e s s u r e . K c = e q u i l i b r i u m c o n s t a n t i n t e r m s o f m o l a r c o n c e n t r a t i o n . R = g a s c o n s t a n t ( 0 . 0 8 2 1 L . a t m m o l . K ) T = t e m p e r a t u r e Δ n = n u m b e r o f m o l e s o f p r o d u c t − n u m b e r o f m o l e s o f r e a c t a n t . \begin{array}{l}\\{{\rm{K}}_{\rm{p}}}\,{\rm{ = }}\,{{\rm{K}}_{\rm{c}}}{\left( {{\rm{RT}}} \right)^{{\rm{\Delta n}}}}\\\\{{\rm{K}}_{\rm{p}}}\,{\rm{ = }}\,\,{\rm{equilibrium}}\,\,{\rm{constant}}\,\,{\rm{in}}\,{\rm{terms}}\,\,{\rm{of}}\,{\rm{partial}}\,{\rm{pressure}}{\rm{.}}\\\\{{\rm{K}}_{\rm{c}}}\,{\rm{ = }}\,\,{\rm{equilibrium}}\,\,{\rm{constant}}\,\,{\rm{in}}\,{\rm{terms}}\,\,{\rm{of}}\,{\rm{molar}}\,\,{\rm{concentration}}{\rm{.}}\\\\{\rm{R}}\,\,\,{\rm{ = }}\,\,{\rm{gas}}\,\,{\rm{constant}}\,\,\left( {{\rm{0}}{\rm{.0821}}\frac{{{\rm{L}}{\rm{.atm}}}}{{{\rm{mol}}{\rm{.K}}}}} \right)\\\\{\rm{T}}\,\,\,{\rm{ = }}\,\,{\rm{temperature}}\\\\{\rm{\Delta n = }}\,\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,{\rm{product}}\,{\rm{ - }}\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,{\rm{reactant}}{\rm{.}}\\\end{array} K p = K c ( R T ) Δ n K p = e q u i l i b r i u m c o n s t a n t i n t e r m s o f p a r t i a l p r e s s u r e . K c = e q u i l i b r i u m c o n s t a n t i n t e r m s o f m o l a r c o n c e n t r a t i o n . R = g a s c o n s t a n t ( 0 . 0 8 2 1 m o l . K L . a t m ) T = t e m p e r a t u r e Δ n = n u m b e r o f m o l e s o f p r o d u c t − n u m b e r o f m o l e s o f r e a c t a n t .
The reaction quotient is represented as Q {\rm{Q}} Q which is used to predict the respective amount of products and reactants. The reaction quotient Q {\rm{Q}} Q of a reaction is measured at a specific point.
Relationship between K and Q:
The reaction at equilibrium is measured byK {\rm{K}} K . The reaction that is not present in equilibrium is represented asQ {\rm{Q}} Q .
In terms of molar concentration, Q {\rm{Q}} Q is represented as,
Q c = [ C ] c [ D ] d [ A ] a [ B ] b {{\rm{Q}}_{\rm{c}}}\,\,{\rm{ = }}\,\,\frac{{{{\left[ {\rm{C}} \right]}^{\rm{c}}}{{\left[ {\rm{D}} \right]}^{\rm{d}}}}}{{{{\left[ {\rm{A}} \right]}^{\rm{a}}}{{\left[ {\rm{B}} \right]}^{\rm{b}}}}} Q c = [ A ] a [ B ] b [ C ] c [ D ] d
In terms of partial pressure, Q {\rm{Q}} Q is represented as,
Q P = ( C ) c ( D ) d ( A ) a ( B ) b {{\rm{Q}}_{\rm{P}}}\,{\rm{ = }}\,\,\frac{{{{\left( {\rm{C}} \right)}^{\rm{c}}}{{\left( {\rm{D}} \right)}^{\rm{d}}}}}{{{{\left( {\rm{A}} \right)}^{\rm{a}}}{{\left( {\rm{B}} \right)}^{\rm{b}}}}} Q P = ( A ) a ( B ) b ( C ) c ( D ) d
IfQ > K {\rm{Q}}\,\,{\rm{ > }}\,\,{\rm{K}} Q > K , the reaction is reactant favored.
IfQ > K {\rm{Q}}\,\,{\rm{ > }}\,\,{\rm{K}} Q > K , the reaction favors products.
If Q = K {\rm{Q}}\,\, = \,\,{\rm{K}} Q = K , the reaction is at equilibrium.
Δ n = m o l e s o f p r o d u c t s − m o l e s o f r e a c t a n t s Δ n = 1 − ( 1 + 1 ) Δ n = − 1 C a l c u l a t i o n f o r d e t e r m i n i n g K c : K p = K c ( R T ) Δ n 1 . 0 0 = K c ( 0 . 0 8 2 1 L . a t m m o l . K × 3 0 0 K ) − 1 \begin{array}{l}\\{\rm{\Delta n}}\,\,{\rm{ = }}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{products}}\,\,{\rm{ - }}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{reactants}}\\\\{\rm{\Delta n}}\,\,{\rm{ = }}\,\,{\rm{1 - (1 + 1)}}\\\\{\rm{\Delta n}}\,\,{\rm{ = }}\,\,{\rm{ - 1}}\\\\{\rm{Calculation}}\,\,{\rm{for}}\,\,{\rm{determining}}\,\,{\rm{Kc:}}\\\\\,\,\,\,{{\rm{K}}_{\rm{p}}}{\rm{ = }}\,\,{{\rm{K}}_{\rm{c}}}{{\rm{(RT)}}^{{\rm{\Delta n}}}}\\\\{\rm{1}}{\rm{.00}}\,\,\,{\rm{ = }}\,\,{{\rm{K}}_{\rm{c}}}\,{\left( {{\rm{0}}{\rm{.0821}}\,\frac{{{\rm{L}}{\rm{.atm}}}}{{{\rm{mol}}{\rm{.K}}}}\,\,{\rm{ \times }}\,\,{\rm{300K}}} \right)^{{\rm{ - 1}}}}\\\end{array} Δ n = m o l e s o f p r o d u c t s − m o l e s o f r e a c t a n t s Δ n = 1 − ( 1 + 1 ) Δ n = − 1 C a l c u l a t i o n f o r d e t e r m i n i n g K c : K p = K c ( R T ) Δ n 1 . 0 0 = K c ( 0 . 0 8 2 1 m o l . K L . a t m × 3 0 0 K ) − 1
1 . 0 0 = K c ( 0 . 0 8 2 1 L . a t m m o l . K × 3 0 0 K ) K c = 1 . 0 0 ( 0 . 0 8 2 1 L . a t m m o l . K × 3 0 0 K ) K c = 2 4 . 6 3 \begin{array}{l}\\{\rm{1}}{\rm{.00}}\,\,\,{\rm{ = }}\,\,\frac{{{{\rm{K}}_{\rm{c}}}}}{{{{\left( {{\rm{0}}{\rm{.0821}}\,\frac{{{\rm{L}}{\rm{.atm}}}}{{{\rm{mol}}{\rm{.K}}}}\,\,{\rm{ \times }}\,\,{\rm{300K}}} \right)}^{}}}}\\\\\,\,\,\,{{\rm{K}}_{\rm{c}}}\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.00}}\,\,\left( {{\rm{0}}{\rm{.0821}}\,\frac{{{\rm{L}}{\rm{.atm}}}}{{{\rm{mol}}{\rm{.K}}}}\,\,{\rm{ \times }}\,\,{\rm{300K}}} \right)\\\\\,\,\,{{\rm{K}}_{\rm{c}}}\,\,{\rm{ = }}\,\,\,{\rm{24}}{\rm{.63}}\\\end{array} 1 . 0 0 = ( 0 . 0 8 2 1 m o l . K L . a t m × 3 0 0 K ) K c K c = 1 . 0 0 ( 0 . 0 8 2 1 m o l . K L . a t m × 3 0 0 K ) K c = 2 4 . 6 3
Part 1
G i v e n : [ X ] = 1 . 0 M [ Y ] = 1 . 0 M [ Z ] = 1 . 0 M \begin{array}{l}\\{\rm{Given:}}\\\\\left[ {\rm{X}} \right]\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.0M}}\\\\\left[ {\rm{Y}} \right]\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.0M}}\\\\\,\left[ {\rm{Z}} \right]\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.0M}}\\\end{array} G i v e n : [ X ] = 1 . 0 M [ Y ] = 1 . 0 M [ Z ] = 1 . 0 M
C a l c u l a t i o n : Q c = [ Z ] [ X ] [ Y ] = 1 1 × 1 Q c = 1 \begin{array}{l}\\{\rm{Calculation:}}\\\\{{\rm{Q}}_{\rm{c}}}\,\,{\rm{ = }}\,\,\frac{{\left[ {\rm{Z}} \right]}}{{\left[ {\rm{X}} \right]\left[ {\rm{Y}} \right]}}\\\\\,\,\,\,\,\,\,{\rm{ = }}\,\,\,\frac{{\rm{1}}}{{{\rm{1}}\,\,{\rm{ \times }}\,\,{\rm{1}}}}\\\\{{\rm{Q}}_{\rm{c}}}\,\,{\rm{ = }}\,\,{\rm{1}}\\\end{array} C a l c u l a t i o n : Q c = [ X ] [ Y ] [ Z ] = 1 × 1 1 Q c = 1
From the Step 1 calculation, the value of K c i s 2 4 . 6 3 {\rm{Kc}}\,\,{\rm{is}}\,\,{\rm{24}}{\rm{.63}} K c i s 2 4 . 6 3 . Therefore, Q c {{\rm{Q}}_{\rm{c}}} Q c is less thanK c {\rm{Kc}} K c .
a)The net reaction will not go to the left because the value of Q c {{\rm{Q}}_{\rm{c}}} Q c is not greater thanK c {\rm{Kc}} K c .
b)The reaction is not in equilibrium because the value of Q c {{\rm{Q}}_{\rm{c}}} Q c is not equal toK c {\rm{Kc}} K c .
c)The reaction will go to the right because the value of Q c {{\rm{Q}}_{\rm{c}}} Q c is less thanK c {\rm{Kc}} K c .
Part 2
G i v e n : P X = 1 . 0 a t m P Z = 1 . 0 a t m P Y = 0 . 5 0 a t m \begin{array}{l}\\{\rm{Given:}}\\\\{{\rm{P}}_{\rm{X}}}{\rm{ = }}\,\,{\rm{1}}{\rm{.0}}\,{\rm{atm}}\\\\{{\rm{P}}_{\rm{Z}}}\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.0}}\,{\rm{atm}}\\\\{{\rm{P}}_{\rm{Y}}}\,\,{\rm{ = }}\,\,{\rm{0}}{\rm{.50}}\,{\rm{atm}}\\\end{array} G i v e n : P X = 1 . 0 a t m P Z = 1 . 0 a t m P Y = 0 . 5 0 a t m
C a l c u l a t i o n o f Q p : Q p = P Z P X P Y = 1 . 0 a t m 1 . 0 a t m × 0 . 5 a t m Q p = 2 . 0 \begin{array}{l}\\{\rm{Calculation}}\,\,{\rm{of}}\,\,{{\rm{Q}}_{\rm{p}}}{\rm{:}}\\\\{{\rm{Q}}_{\rm{p}}}\,{\rm{ = }}\,\,\frac{{{{\rm{P}}_{\rm{Z}}}}}{{{{\rm{P}}_{\rm{X}}}\,{{\rm{P}}_{\rm{Y}}}}}\,\,\\\\\,\,\,\,\,\,\,{\rm{ = }}\,\,\frac{{{\rm{1}}{\rm{.0}}\,{\rm{atm}}\,}}{{{\rm{1}}{\rm{.0}}\,{\rm{atm}}\,{\rm{ \times }}\,{\rm{0}}{\rm{.5}}\,{\rm{atm}}\,}}\\\\{{\rm{Q}}_{\rm{p}}}\,{\rm{ = }}\,\,\,{\rm{2}}{\rm{.0}}\\\end{array} C a l c u l a t i o n o f Q p : Q p = P X P Y P Z = 1 . 0 a t m × 0 . 5 a t m 1 . 0 a t m Q p = 2 . 0
From the given data, the value of K p i s 1 . 0 0 {\rm{Kp}}\,\,{\rm{is}}\,\,{\rm{1}}{\rm{.00}} K p i s 1 . 0 0 . Therefore, Q p {{\rm{Q}}_{\rm{p}}} Q p is greater thanK p {\rm{Kp}} K p .
c)The net reaction will not go to the right because the value of Q p {{\rm{Q}}_{\rm{p}}} Q p is not less thanK p {\rm{Kp}} K p .
d)The reaction is not in equilibrium because the value of Q p {{\rm{Q}}_{\rm{p}}} Q p is not less thanK p {\rm{Kp}} K p .
a)The reaction will go to the left because the value of Q p {{\rm{Q}}_{\rm{p}}} Q p is greater thanK p {\rm{Kp}} K p .
Ans: Part 1
b)Net reaction goes to the right.
Part 2
a)Net reaction goes to the left.