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For the reaction

$$\ce{1/2 N2(g) + 3/2 H2(g) -> NH3(g)}$$

$ΔG$ at $\pu{298.15 K}$ is $\pu{-3980 cal},$ and the change in enthalpy is given by the expression:

$$ΔH = -9.19 - 7.12T + 3.182\cdot 10^{-3}T^2 - 2.64\cdot 10^{-7}T^3$$

Calculate $ΔG$ at $\pu{1000 K}$ from these data. Note: use the expression

$$\left[\frac{∂}{∂T} \frac{ΔG}{T}\right] = \frac{-ΔH}{T^2}$$

integrate the resulting expression of the $ΔH$ given.

This is what I understood so far. Using the given expression of the partial derivative I integrated on both sides:

$$∫\left[\frac{∂}{∂T} \frac{ΔG}{T}\right] = - ∫\frac{ΔH}{T^2}$$

I put the given expression of $ΔH$ on the right side of the integral.

$$∫\left[\frac{∂}{∂T} \frac{ΔG}{T}\right] = - ∫\frac{-9.19 - 7.12T + 3.182\cdot 10^{-3}T^2 - 2.64\cdot 10^{-7}T^3}{T^2}$$

Eliminate the $T$ where I could:

$$∫\left[\frac{∂}{∂T} \frac{ΔG}{T}\right] = 9.19 ∫\frac{1}{T^2}\mathrm dT + 7.12 ∫\frac{1}{T}\mathrm dT - 3.182\cdot 10^{-3}∫\mathrm dT + 2.64\cdot 10^{-7} T^3∫T\mathrm dT$$

$$\frac{ΔG}{T} + C = -\frac{9.19}{T} + 7.12\ln T - 3.182\cdot 10^{-3}T + 1.32\cdot 10^{-7}T^2$$

To get $ΔG$ I evaluated the final expression from $298.15$ to $\pu{1000 K}$ and got that

$$\frac{ΔG}{T} + C = 126.6707439$$

  1. Since the expression from the integral is all temperatures, why I don't get units of energy instead? Am I missing something during the integration?

  2. If I am not missing anything, the $126.6707439$, whatever unit it has, is that the $ΔG$ at $\pu{1000 K}$ I am supposed to get?

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  • 2
    $\begingroup$ You integrated indefinitely on L.H.S. but took limits on L.H.S.. Either integrate indefinitely on both sides and put values at 298.15K or take limits on both sides. Everything is correct till the second last equation $\endgroup$ – Groverkss May 16 at 3:19
  • $\begingroup$ It seems to me the initial equation implicitly supposes dS/dT=0. But, it does not seem right to me, as integral dQ/T up the temperature for reagents and the same downwards for products do not seem to me as zero, if sumed up. $\endgroup$ – Poutnik May 16 at 8:00
  • $\begingroup$ @Poutnik, this is the Gibbs-Helmholtz equation and assumes $p$ is constant. Entropy is given by $\displaystyle \left( \frac{\partial G}{\partial T}\right)_p=-S$ and starting with $dG=VdP-SdT$ and $G=H-TS$ and after a few steps $\displaystyle \left (\frac{\partial G}{\partial T}\right)_p=\frac{G-H}{T}$ results which is the same as the equation in the question $\endgroup$ – porphyrin May 16 at 14:51
  • $\begingroup$ Integrating the left had side of the equation wrt $T$ gives $\displaystyle \frac{\Delta G_{T_2}}{ T_2}- \frac{\Delta G_{T_1}}{ T_1}$ $\endgroup$ – porphyrin May 16 at 15:13

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