In two experiments, the amount of ammonia formed is recorded as a function of time. $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92\,\text{kJ mol}^{-1}$ For experiment 1, $3\,\text{mol}$ of $\text{H}_2(g)$ and $1\,\text{mol}$ of $\text{N}_2(g)$ are allowed to react at $45\,^{\circ}\text{C}$ and under a pressure of $200\,\text{atm}$. A graph of ammonia volume produced against time is drawn. Experiment 2 is then carried out. It is the same as experiment 1 except for one condition. What is different in experiment 2 from experiment 1?
- AAn iron catalyst is present in experiment 2.
- B$2\,\text{mol}$ of helium gas is present in the reaction mixture in experiment 2.
- CA pressure of $250\,\text{atm}$ is used in experiment 2.
- DA temperature of $600\,^{\circ}\text{C}$ is used in experiment 2.