Question

In which direction will the net reaction proceed.
X(g) + Y(g) <==> Z(g) .. Kp = 1.00 at 300k
for each of these sets of initial conditions?

1) [X] = [Y] = [Z] = 1.0 M
a] net reaction goes to the left [this one?]
b] net reaction goes to the right
c] reaction is at equilibrium

2. Px = Pz = 1.0 atm, Py = 0.50 atm

a] net reaction goes to the left
b] net reaction goes to the right
c] reaction is at equilibrium

Answer 1

Concepts and reason

Equilibrium constant is the ratio between the concentration of products to the concentration of reactants. The equilibrium constant is expressed as${\rm{K}}$. If the value of ${\rm{K}}$is more than 1, the equilibrium favors products and if the value of ${\rm{K}}$is less than 1, the equilibrium favors reactants.

Fundamentals

For a simple equilibrium reaction,

The equilibrium constant with respect to molar concentration is represented as${\rm{Kc}}$.

${\rm{Kc}}\,\,{\rm{ = }}\,\,\frac{{{{\left[ {\rm{C}} \right]}^{\rm{c}}}{{\left[ {\rm{D}} \right]}^{\rm{d}}}}}{{{{\left[ {\rm{A}} \right]}^{\rm{a}}}{{\left[ {\rm{B}} \right]}^{\rm{b}}}}}$

The equilibrium constant in terms of partial pressure can be represented as${{\rm{K}}_{\rm{p}}}$.

${{\rm{K}}_{\rm{P}}}\,{\rm{ = }}\,\,\frac{{{{\left( {\rm{C}} \right)}^{\rm{c}}}{{\left( {\rm{D}} \right)}^{\rm{d}}}}}{{{{\left( {\rm{A}} \right)}^{\rm{a}}}{{\left( {\rm{B}} \right)}^{\rm{b}}}}}$

The relation between both ${\rm{Kc}}$and ${{\rm{K}}_{\rm{p}}}$is:

$\begin{array}{l}\\{{\rm{K}}_{\rm{p}}}\,{\rm{ = }}\,{{\rm{K}}_{\rm{c}}}{\left( {{\rm{RT}}} \right)^{{\rm{\Delta n}}}}\\\\{{\rm{K}}_{\rm{p}}}\,{\rm{ = }}\,\,{\rm{equilibrium}}\,\,{\rm{constant}}\,\,{\rm{in}}\,{\rm{terms}}\,\,{\rm{of}}\,{\rm{partial}}\,{\rm{pressure}}{\rm{.}}\\\\{{\rm{K}}_{\rm{c}}}\,{\rm{ = }}\,\,{\rm{equilibrium}}\,\,{\rm{constant}}\,\,{\rm{in}}\,{\rm{terms}}\,\,{\rm{of}}\,{\rm{molar}}\,\,{\rm{concentration}}{\rm{.}}\\\\{\rm{R}}\,\,\,{\rm{ = }}\,\,{\rm{gas}}\,\,{\rm{constant}}\,\,\left( {{\rm{0}}{\rm{.0821}}\frac{{{\rm{L}}{\rm{.atm}}}}{{{\rm{mol}}{\rm{.K}}}}} \right)\\\\{\rm{T}}\,\,\,{\rm{ = }}\,\,{\rm{temperature}}\\\\{\rm{\Delta n = }}\,\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,{\rm{product}}\,{\rm{ - }}\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,{\rm{reactant}}{\rm{.}}\\\end{array}$

The reaction quotient is represented as ${\rm{Q}}$which is used to predict the respective amount of products and reactants. The reaction quotient ${\rm{Q}}$of a reaction is measured at a specific point.

Relationship between K and Q:

The reaction at equilibrium is measured by${\rm{K}}$. The reaction that is not present in equilibrium is represented as${\rm{Q}}$.

In terms of molar concentration, ${\rm{Q}}$is represented as,

${{\rm{Q}}_{\rm{c}}}\,\,{\rm{ = }}\,\,\frac{{{{\left[ {\rm{C}} \right]}^{\rm{c}}}{{\left[ {\rm{D}} \right]}^{\rm{d}}}}}{{{{\left[ {\rm{A}} \right]}^{\rm{a}}}{{\left[ {\rm{B}} \right]}^{\rm{b}}}}}$

In terms of partial pressure, ${\rm{Q}}$is represented as,

${{\rm{Q}}_{\rm{P}}}\,{\rm{ = }}\,\,\frac{{{{\left( {\rm{C}} \right)}^{\rm{c}}}{{\left( {\rm{D}} \right)}^{\rm{d}}}}}{{{{\left( {\rm{A}} \right)}^{\rm{a}}}{{\left( {\rm{B}} \right)}^{\rm{b}}}}}$

If${\rm{Q}}\,\,{\rm{ > }}\,\,{\rm{K}}$, the reaction is reactant favored.

If${\rm{Q}}\,\,{\rm{ > }}\,\,{\rm{K}}$, the reaction favors products.

If ${\rm{Q}}\,\, = \,\,{\rm{K}}$, the reaction is at equilibrium.

$\begin{array}{l}\\{\rm{\Delta n}}\,\,{\rm{ = }}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{products}}\,\,{\rm{ - }}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{reactants}}\\\\{\rm{\Delta n}}\,\,{\rm{ = }}\,\,{\rm{1 - (1 + 1)}}\\\\{\rm{\Delta n}}\,\,{\rm{ = }}\,\,{\rm{ - 1}}\\\\{\rm{Calculation}}\,\,{\rm{for}}\,\,{\rm{determining}}\,\,{\rm{Kc:}}\\\\\,\,\,\,{{\rm{K}}_{\rm{p}}}{\rm{ = }}\,\,{{\rm{K}}_{\rm{c}}}{{\rm{(RT)}}^{{\rm{\Delta n}}}}\\\\{\rm{1}}{\rm{.00}}\,\,\,{\rm{ = }}\,\,{{\rm{K}}_{\rm{c}}}\,{\left( {{\rm{0}}{\rm{.0821}}\,\frac{{{\rm{L}}{\rm{.atm}}}}{{{\rm{mol}}{\rm{.K}}}}\,\,{\rm{ \times }}\,\,{\rm{300K}}} \right)^{{\rm{ - 1}}}}\\\end{array}$

$\begin{array}{l}\\{\rm{1}}{\rm{.00}}\,\,\,{\rm{ = }}\,\,\frac{{{{\rm{K}}_{\rm{c}}}}}{{{{\left( {{\rm{0}}{\rm{.0821}}\,\frac{{{\rm{L}}{\rm{.atm}}}}{{{\rm{mol}}{\rm{.K}}}}\,\,{\rm{ \times }}\,\,{\rm{300K}}} \right)}^{}}}}\\\\\,\,\,\,{{\rm{K}}_{\rm{c}}}\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.00}}\,\,\left( {{\rm{0}}{\rm{.0821}}\,\frac{{{\rm{L}}{\rm{.atm}}}}{{{\rm{mol}}{\rm{.K}}}}\,\,{\rm{ \times }}\,\,{\rm{300K}}} \right)\\\\\,\,\,{{\rm{K}}_{\rm{c}}}\,\,{\rm{ = }}\,\,\,{\rm{24}}{\rm{.63}}\\\end{array}$

Part 1

$\begin{array}{l}\\{\rm{Given:}}\\\\\left[ {\rm{X}} \right]\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.0M}}\\\\\left[ {\rm{Y}} \right]\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.0M}}\\\\\,\left[ {\rm{Z}} \right]\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.0M}}\\\end{array}$

$\begin{array}{l}\\{\rm{Calculation:}}\\\\{{\rm{Q}}_{\rm{c}}}\,\,{\rm{ = }}\,\,\frac{{\left[ {\rm{Z}} \right]}}{{\left[ {\rm{X}} \right]\left[ {\rm{Y}} \right]}}\\\\\,\,\,\,\,\,\,{\rm{ = }}\,\,\,\frac{{\rm{1}}}{{{\rm{1}}\,\,{\rm{ \times }}\,\,{\rm{1}}}}\\\\{{\rm{Q}}_{\rm{c}}}\,\,{\rm{ = }}\,\,{\rm{1}}\\\end{array}$

From the Step 1 calculation, the value of ${\rm{Kc}}\,\,{\rm{is}}\,\,{\rm{24}}{\rm{.63}}$. Therefore, ${{\rm{Q}}_{\rm{c}}}$is less than${\rm{Kc}}$.

a)The net reaction will not go to the left because the value of ${{\rm{Q}}_{\rm{c}}}$is not greater than${\rm{Kc}}$.

b)The reaction is not in equilibrium because the value of ${{\rm{Q}}_{\rm{c}}}$is not equal to${\rm{Kc}}$.

c)The reaction will go to the right because the value of ${{\rm{Q}}_{\rm{c}}}$is less than${\rm{Kc}}$.

Part 2

$\begin{array}{l}\\{\rm{Given:}}\\\\{{\rm{P}}_{\rm{X}}}{\rm{ = }}\,\,{\rm{1}}{\rm{.0}}\,{\rm{atm}}\\\\{{\rm{P}}_{\rm{Z}}}\,\,{\rm{ = }}\,\,{\rm{1}}{\rm{.0}}\,{\rm{atm}}\\\\{{\rm{P}}_{\rm{Y}}}\,\,{\rm{ = }}\,\,{\rm{0}}{\rm{.50}}\,{\rm{atm}}\\\end{array}$

$\begin{array}{l}\\{\rm{Calculation}}\,\,{\rm{of}}\,\,{{\rm{Q}}_{\rm{p}}}{\rm{:}}\\\\{{\rm{Q}}_{\rm{p}}}\,{\rm{ = }}\,\,\frac{{{{\rm{P}}_{\rm{Z}}}}}{{{{\rm{P}}_{\rm{X}}}\,{{\rm{P}}_{\rm{Y}}}}}\,\,\\\\\,\,\,\,\,\,\,{\rm{ = }}\,\,\frac{{{\rm{1}}{\rm{.0}}\,{\rm{atm}}\,}}{{{\rm{1}}{\rm{.0}}\,{\rm{atm}}\,{\rm{ \times }}\,{\rm{0}}{\rm{.5}}\,{\rm{atm}}\,}}\\\\{{\rm{Q}}_{\rm{p}}}\,{\rm{ = }}\,\,\,{\rm{2}}{\rm{.0}}\\\end{array}$

From the given data, the value of ${\rm{Kp}}\,\,{\rm{is}}\,\,{\rm{1}}{\rm{.00}}$. Therefore, ${{\rm{Q}}_{\rm{p}}}$is greater than${\rm{Kp}}$.

c)The net reaction will not go to the right because the value of ${{\rm{Q}}_{\rm{p}}}$is not less than${\rm{Kp}}$.

d)The reaction is not in equilibrium because the value of ${{\rm{Q}}_{\rm{p}}}$is not less than${\rm{Kp}}$.

a)The reaction will go to the left because the value of ${{\rm{Q}}_{\rm{p}}}$is greater than${\rm{Kp}}$.

Ans: Part 1

b)Net reaction goes to the right.

Part 2

a)Net reaction goes to the left.