How to calculate the variation of the standard reaction enthalpy for the CO, CO2, H2O, H2 equilibrium?

For the reaction $$\ce{CO2 (g) + H2 (g) <=> CO (g) + H2O (g)}$$ the equilibrium constant $$K_p$$ has a value of $$0.534$$ at $$\pu{960 K}$$ and $$1.571$$ at $$\pu{1260 K}$$. Calculate the variation of the standard reaction enthalpy, assumed constant in the temperature range considered. Also calculate the $$K_p$$ value of the above reaction at standard conditions at $$\pu{25 ^\circ C}$$.

My procedure:

$$K_c = K_p(RT)^{\Delta \nu (g)} = K_p(RT)^0 = K_p \cdot 1 = K_p$$

From the van't Hoff equation:

\begin{align} \ln K_2 - \ln K_1 &= -\frac{\Delta H^\circ_\mathrm{r}}{R} \left[\frac{1}{T_2} - \frac{1}{T_1}\right]\\ \Delta H^\circ_\mathrm{r} &= \pu{+36.177 kJ//mol} \end{align}

Now, for $$K_p$$ at $$\pu{25 ^\circ C}$$: $$K_p = \exp\left\{-\frac{\Delta G^\circ_\mathrm{r}}{RT}\right\}$$

Which of these two equations should I use to calculate $$\Delta G^\circ_\mathrm{r}$$? \begin{align} \Delta G^\circ_\mathrm{r} &= \Delta H^\circ_\mathrm{r} - T\Delta S^\circ_\mathrm{r}\tag1\\ \Delta G^\circ_\mathrm {r} &= \sum_\text{products} \nu\Delta G^\circ_\mathrm{f} - \sum_\text{reagents} \nu\Delta G^\circ_\mathrm{f} \tag2 \end{align}

• There are a couple of problems with your post: (1) a missing value of Kp at 1571 K (2) an unspecified state for H2 (a gas or ?). For the answer, I think just use 298 K in the van't Hoff expression together to compute Kp at the new temperature. No need to compute $\Delta G^\circ$ – Buck Thorn Jun 2 '19 at 8:15
• The post has been edited several times by other users. Now I have corrected it – user3204810 Jun 2 '19 at 8:24
• 1.571 is K_2, T_2 is 1260 Kelvin – user3204810 Jun 2 '19 at 8:49
• Just plug in $\Delta H$, T1, T2 (298 K) and K1 into the van't Hoff expression and solve for K2. – Buck Thorn Jun 2 '19 at 8:59
• @mhchem Unfortunately the typography you have used was completely wrong in this case, as p is for pressure, which should be italic. – Martin - マーチン Jun 3 '19 at 15:22