To calculate $\Delta H$, the change in enthalpy at $\mathrm{100^\circ C}$ for the reaction below, one needs what addition information?

$$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$$

$\Delta H^\circ = \pu{-92.0 kJ}$ at $\mathrm{25^\circ C}$

(a) The equilibrium constant for the reaction at $\mathrm{100^\circ C}$

(b) The molar heat capacities of the reaction and the products as a function of temperature

(c) $\Delta E^\circ$, the standard internal energy change for the reaction

(d) The partial pressures of the reactants and products at $\mathrm{100^\circ C}$

(e) The entropies of formation for the reactants and products at $\mathrm{100^\circ C}$

The correct answer is (b), but I'm not entirely sure why.

I know it can't be (a) because I think we need $\Delta S$ as well. Also, (c) won't be helpful because it is at standard temperature, and we are wanting information at $\mathrm{100^\circ C}$. Finally, (e), seems correct but they are actually talking about entropy instead of enthalpy. However, I'm not sure how to rule out (d) or justify (b).


2 Answers 2


@Chet already gave the qualitative answer but the actual equation it is based on is:

$$ \Delta H^\circ(T)=\Delta H{^\circ}_{T_1} + \int_{T_1}^{T} \Delta C_p{^\circ}(T^\prime ) \,\mathrm{d}T^\prime$$

If you know the enthalpy at a reference temperature and the heat capacity as a function of temperature you can calculate the enthalpy change at any other temperature.

Actually d) and e) together are not wrong. We all know these equations: $$\Delta G= - RT\ln (K) \qquad \Delta G= \Delta H - T\Delta S $$

With d) you can calculate $\Delta G$ and with e) you can calculate $\Delta H$.

Lets look deeper into: $\Delta G^\circ$, $\Delta G$, $K$ and $Q$. To answer your second question completely I would recommend to look again into a textbook of your choice. Especially the definitions of activity and chemical potential.

If you know the chemical potential $\mu^\circ$ of a pure substance you can calculate the chemical potential of mixtures if you know the activities: $$\mu= \mu^\circ + RT \ln(a)$$

Now we make two assumptions:

  1. We have constant temperature and pressure. ($\mu = \Delta G$)
  2. We can calculate the activities simply over the partial pressures or concentrations.

This leads then to your equation: $$ \Delta G= \Delta G^\circ + RT \ln(Q) $$ where $Q$ is the reaction quotient. Since we assume equilibrium, we know that $\Delta G=0$ and $Q = K$. This leads to: $$ 0= \Delta G^\circ + RT \ln(K) \Rightarrow \Delta G^\circ = -RT \ln(K) $$

The last equation is the handy one you use for calculating equilibrium constants from energies and vice versa. So to be really precise one has to write: $$ \Delta G^\circ= \Delta H^\circ - T\Delta S^\circ $$

But one usually skips the superscript $^\circ$. At equilibrium $Q$ and $K$ are essentially the same.

  • 1
    $\begingroup$ Forgive if I'm wrong, but isn't ∆G˚ = -RTln(K) and ∆G = ∆G˚ + RTln(Q)? So, how can one get ∆G if we only have K? $\endgroup$
    – coloratura
    Sep 3, 2015 at 22:46
  • $\begingroup$ I edited my answer to adress this question. Perhaps you edit your question. $\endgroup$
    – mcocdawc
    Sep 4, 2015 at 9:38

Since $\Delta H$ is a function of state, you can get $\Delta H$ at 100 C by taking the reactants at 100 C, cooling them down to 25 C, then carrying out the reaction at 25 C, and then heating the products up to 100 C. This procedure makes use of Hess' Law. Getting the amount of heat removed to take the reactants from 100 C to 25 C requires using the molar heat capacities. Same for heating the products back up to 100 C.


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