Exit to H2 Chemistry (9476): Chemical Equilibria

Haber Process (Case Study) (A Level) Quiz

Derived lesson quiz from the Chemical Equilibria bank focused on Haber-process yield-rate compromises and operating conditions.

GCE A-Level H2 Chemistry 9476-2027

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A catalyst is added to a reversible reaction already at equilibrium, with temperature kept constant. What changes?

Shifts equilibrium to the right
Shifts equilibrium to the left
Changes the value of $K$
Both reaction rates increase equally; the equilibrium composition is unchanged

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), what happens when the equilibrium mixture is compressed by reducing its volume at constant temperature?

Equilibrium shifts left; less $\ce{NH3}$ forms
Equilibrium shifts right; more $\ce{NH3}$ forms
No change because pressure does not affect equilibria
The value of $K_c$ increases

For an exothermic forward reaction, what is the effect of increasing temperature on the equilibrium position?

Shifts to products (right)
Shifts to reactants (left)
No change because temperature only affects rate
Depends only on the number of moles of gas

Which change can alter the value of an equilibrium constant $K$ for a given reaction?

Changing concentration
Changing pressure
Changing temperature
Adding a catalyst

Which statement about Le Chatelier’s principle is correct?

The equilibrium always shifts to form more products
The equilibrium always shifts to the side with fewer moles of gas
The system shifts to oppose the change imposed
The value of $K$ changes to oppose the change imposed

Which set of conditions is a typical industrial compromise for the Haber process?

Very low temperature, very low pressure, no catalyst
High temperature, low pressure, no catalyst
Moderate temperature, high pressure, iron catalyst
Moderate temperature, low pressure, iron catalyst

Why is a very low temperature not used in the Haber process even though it increases equilibrium yield?

Because it makes the reaction endothermic
Because the rate becomes too slow for industrial production
Because it decreases the value of $K$
Because catalysts only work at high temperatures

What is the effect of continuously removing $\ce{NH3}$ as it forms in the Haber process?

Shifts equilibrium left, reducing yield
Shifts equilibrium right, increasing overall yield
Increases $K_c$ at constant temperature
Makes the reaction irreversible

If $K$ increases when temperature increases, what does this suggest about the forward reaction?

Forward reaction is exothermic
Forward reaction is endothermic
Forward reaction must involve gases only
Forward reaction must have fewer moles of gas on the right

Haber Process (Case Study) (A Level) Quiz

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