Question

HCN is produced commercially by the exothermic reaction                             

2NH₃(g)    +   3O₂(g)    +   2CH₄(g)    →    2HCN(g)    +   6H₂O(g)

The reaction is performed at 1000^oC in the presence of a catalyst.  Is the high temperature being used to speed up the reaction or instead to make the reaction more favorable?  Explain your choice.  Look at the signs of ∆H and ∆S for the reaction.  

For HCN(g)         ΔH_f= 135 kJ/mole       S= 202 J/mole^.K      ΔG_f= 125 kJ/mole

Answer 1

As the given reaction is exothermic which means the sign of H for the reaction is negative.

From the equation, the number of gaseous moles in the product side (8 moles = 2 moles HCN and 6 moles H2O) is higher than that of reactant side (7 moles = 2 moles NH3 + 3 moles O2 + 2 moles CH4). As the gaseous molecules increase from reactant to product side, the entropy increases. As the entropy is defined as the randomness of the system. So, the sign of S for the reaction is positive.

Now, from the thermodynamic equation,

G = H - TS = (-) - (+)(+) = (-)

So, the reaction is spontaneous at all the temperatures. As the temperature increases the sign of G become more and more negative which means reaction become more favorable.

Hence, the high temperature is being used to make the reaction more favorable.

Note: The catalyst lower the activation energy of the reaction and speed up the reaction. So, catalyst define the kinetic of the reaction while temperature define the thermodynamic of the reaction.