Question

Consider the following equilibrium reaction: 4NH3(g) + 5O2(g) <---> 4NO(g) + 6H2O(g) Heat = -904.4 kJ

How does each of the following changes affect the yield of NO at equilibrium? Answer increase, decrease, or no change.

a. increase [NH3]
b. Increase [H2O]
c. Decrease [O2]
d. Decrease the volume of the container in which the raction occurs.
e. Add a catalyst
f. Increase temperature

Answer 1

Concepts and reason

In a chemical reaction, the chemical equilibrium is a state of system in which the concentration of reactants and products do not change with time.

At equilibrium, the reaction seems to be stopped due to the simultaneous rate change in the concentration of reactants and products.

Equilibrium attains a dynamic situation indicating that: the rate of forward reaction is equal to the rate of backward reaction and there is no net change in concentration.

Fundamentals

Le Chatliers principle:

If a system at equilibrium is disturbed, the equilibrium moves in such a way to counteract the change.

Factors affecting chemical equilibrium are as follows:

oConcentration

oTemperature

oPressure ( or volume)

oFactors Affecting Equilibrium:

(a)

Consider the given equilibrium reaction:

${\rm{4N}}{{\rm{H}}_{\rm{3}}}\left( {\rm{g}} \right)\,{\rm{ + }}\,{\rm{5}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)\,\longrightarrow{{}}\,{\rm{4NO}}\left( {\rm{g}} \right)\,{\rm{ + }}\,{\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right)\,\,\,\,\,\,\,\,\,\,{\rm{Heat}}\,{\rm{ = }}\,{\rm{ - 904}}{\rm{.4kJ}}$

Increase $\left[ {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right]$

${\rm{N}}{{\rm{H}}_{\rm{3}}}$ is the reactant of the given reaction. So, when there is an increase in the concentration of reactant, it causes an increase in the yield of ${\rm{NO}}$ .

(b)

Consider the given equilibrium reaction:

${\rm{4N}}{{\rm{H}}_{\rm{3}}}\left( {\rm{g}} \right)\,{\rm{ + }}\,{\rm{5}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)\,\longrightarrow{{}}\,{\rm{4NO}}\left( {\rm{g}} \right)\,{\rm{ + }}\,{\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right)\,\,\,\,\,\,\,\,\,\,{\rm{Heat}}\,{\rm{ = }}\,{\rm{ - 904}}{\rm{.4kJ}}$

Increase $\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]$

${{\rm{H}}_{\rm{2}}}{\rm{O}}$ is the product of the given reaction. So, when there is an increase in the concentration of product, it causes a decrease in the yield of ${\rm{NO}}$ .

(c)

Consider the given equilibrium reaction:

${\rm{4N}}{{\rm{H}}_{\rm{3}}}\left( {\rm{g}} \right)\,{\rm{ + }}\,{\rm{5}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)\,\longrightarrow{{}}\,{\rm{4NO}}\left( {\rm{g}} \right)\,{\rm{ + }}\,{\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right)\,\,\,\,\,\,\,\,\,\,{\rm{Heat}}\,{\rm{ = }}\,{\rm{ - 904}}{\rm{.4kJ}}$

Decrease $\left[ {{{\rm{O}}_{\rm{2}}}} \right]$

${{\rm{O}}_{\rm{2}}}$ is the reactant of the given reaction. So, when there is a decrease in the concentration of reactant, it causes a decrease in the yield of ${\rm{NO}}$ .

(d)

Consider the given equilibrium reaction:

${\rm{4N}}{{\rm{H}}_{\rm{3}}}\left( {\rm{g}} \right)\,{\rm{ + }}\,{\rm{5}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)\,\longrightarrow{{}}\,{\rm{4NO}}\left( {\rm{g}} \right)\,{\rm{ + }}\,{\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right)\,\,\,\,\,\,\,\,\,\,{\rm{Heat}}\,{\rm{ = }}\,{\rm{ - 904}}{\rm{.4kJ}}$

Decrease the volume of the container in which the reaction occurs.

Change in the volume, effects the yield of the reaction.

Volume is inversely proportional to the pressure.

Decrease in the volume, increases the pressure of the container.

When there is an increase in the pressure of the container, equilibrium shifts to the side in which there are less number of moles. In the reactant side, only 9 moles are present, so the equilibrium shifts to the reactant side.

Therefore, when there is a decrease in the volume, then there is also a decrease in the yield of ${\rm{NO}}$ .

Ans: Part a

Increase the yield of ${\rm{NO}}$

Part b

Decrease the yield of ${\rm{NO}}$

Part c

Decrease the yield of ${\rm{NO}}$

Part d

Decrease the yield of ${\rm{NO}}$