Quick questions on Haber and contact processes explained: HSC Chemistry Module 8

Both processes are reversible, gas-phase, exothermic syntheses, and both face the identical design tension: cooling the system would raise the equilibrium yield, but it would also make the reaction too slow to be commercially useful.

Pressure is chosen by the same Le Chatelier logic, applied to the change in the total number of moles of gas across each reaction.

A catalyst changes ONLY the rate, never the position of equilibrium. Iron in the Haber process and $V_2O_5$ in the contact process each provide an alternative reaction pathway with a lower activation energy, so a larger fraction of molecular collisions succeed at a given temperature. This lets both processes run at their compromise temperature with an economically useful rate, without changing $K_{eq}$ or the equilibrium yield that temperature and pressure alone would produce.

For the Haber process, state the direction of the equilibrium yield shift (or "no change") caused by: (a) increasing pressure; (b) decreasing temperature; (c) adding a catalyst. [3 marks]

Explain, using both Le Chatelier's principle and collision theory, why the contact process is run at 450 degrees C rather than 50 degrees C. [4 marks]

A Haber reactor is fed 4.00 mol of $N_2$ with excess $H_2$, achieving 15.0% single-pass conversion. Calculate the moles, then the mass, of $NH_3$ produced. [3 marks]