Question

What will be the equilibrium constant at `127^circC`, if `K` at `27^circC` is `4` for reaction `N_2+3H_2->2NH_3`; `DeltaH=-46.06kJcolor(white)(l)mol^(-1)`?

Answer 1

We're looking to relate two sets of conditions with a measure activity that is dependent on those conditions. With this equation,

`ln(K_1/K_2) = (DeltaH)/R(1/T_2 - 1/T_1)`

we'll do just that.

A few common mistakes are (i) not using absolute temperature, and (ii) using the wrong gas constant. The new equilibrium constant is,

`ln(4) - ln(K_2) = ((-46.06kJ)/(mol) * (10^3J)/(kJ))/((8.314J)/(mol*K))(1/(400K) - 1/(300K))`
`therefore K_(127°C) approx 4.0*10^-2`

This may seem counterintuitive, since activity is generally directly proportional to temperature in a chemical system. However, the rate forward won't increase as much when the system is exothermic. Hence, this is likely the answer.