Consider the following reaction at equilibrium, which one of these disturbances will make the reverse reaction faster? 2nh3 (g) ↔ n2 (g) + 3h2 (g) δh° =46.1 kj at 298k a increasing the temperature decreasing the concentration of n2 reducing the container volume increasing the partial pressure of nh3

The given process is reverse of the common Haber's process , which is synthesis of ammonia from nitrogen and hydrogen and that is an exothermic reaction

In the given reaction the the enthalpy change is positive hence it is endothermic reaction

2NH3 (g) ↔ N2 (g) + 3H2 (g) ΔH° =46.1 kJ

i) for endothermic reaction if we increase the temperature the forward reaction goes faster hence first option is incorrect

ii) if we decrease the concentration of either product the reaction will again go in forward direction , so the option decreasing concentration of N2 is also wrong

iii) if we reduce the volume of container it means we are increasing the moles / volume, hence reaction will go in the direction where the number of moles of gaseous molecules is less so reaction will go in backward direction

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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