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I'm trying to calculate the $\Delta\mathrm{G}$ of the following reaction. Are my calculations correct? I'm also a bit puzzled, 'cause this is a endoergonic reaction right? But the $\Delta\mathrm{G}$ is negative, so the reaction would be spontaneous if my calculations are correct and if the activation energy is reached right?

CH3OH + H2O --> HCOOH + 4H        Delta H = 99.49 kJ/mol

% Standard enthalpy of formation of every compound

% -238.7 kJ/mol CH3OH
% -285.830 kJ/mol H2O
% -425 kJ/mol HCOOH
%    0 kJ/mol H2
% -393.51 kJ/mol CO2

% Standard entropy of formation of every compound

% 126.8 J/mol*K CH3OH
%  69.91 J/mol*K H2O
% 129 J/mol*K HCOOH
% 130.6 J/mol*K H2
% 213.6 J/mol*K CO2

delta_H_1 = -425 - (-238.7 - 285.830);
delta_S_1 = 129 + 130.6 - (126.8 + 69.91);
T = 273.15 + 20;

delta_G_1 = delta_H_1 - T * delta_S_1;

Correct version

delta_H_1 = -425 - (-238.7 - 285.830);
delta_S_1 = 129 + 4*130.6 - (126.8 + 69.91);
T = 273.15 + 25;

delta_G_1 = delta_H_1 - T * delta_S_1;
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$$ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

Gibbs Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibbs Free energy describes how much "free energy" is available, or how much work is possible.

$\Delta G^\circ$ can either be negative or positive even if $\Delta H^\circ$ is $+$. Therefore, we have to look at all three terms, $\Delta H^\circ$, $T$, and $\Delta S^\circ$

The Gibbs Free Energy, enthalpy, and entropy are state functions. Although I won't derive it here, we can find $\Delta G$, or $\Delta H$, or $\Delta S$, by using Hess's Law.

The information provided above gives us the $\Delta H_{rxn}$ which is $99.49 \text{kJ/mol}$ and thus endothermic.

We calculate entropy from the a list of experimental tabulated values:

$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants} $$

$$\Delta S^\circ_{rxn} = (129.0 \text{J}\cdot\text{mol}^{-1}\text{K} + 4*130.6 \text{J}\cdot\text{mol}^{-1}\text{K}) - (126.8 \text{J}\cdot \text{mol}^{-1}\text{K} + 69.91 \text{J}\cdot\text{mol}^{-1}\text{K})$$

$$\Delta S^\circ_{rxn} = 454.69 \text{J}\cdot\text{mol}^{-1}\text{K} = 0.45469 \text{kJ}\cdot\text{mol}^{-1}\text{K} $$

While solving for entropy, you forgot to multiply the entropy of hydrogen gas with the number of moles. Also, you should always include the states of your reactants and products, as thermodynamic data for compounds differs for different states.

Another thing that is incorrect is standard temperature for thermodynamic reactions is at $25^\circ C$ not $20^\circ C$ and thus your temperature should be $298.15 K$

I'm also a bit puzzled, 'cause this is a endoergonic reaction right? But the ΔG is negative, so the reaction would be spontaneous if my calculations are correct and if the activation energy is reached right?

This is where a lot of people make mistakes. It is true, that in general, endothermic reactions with a positive enthalpy usually have a positive $\Delta G$ and are thus not spontaneous.

However, if we look at our equation:

$$ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

We have to consider our other terms and not only enthalpy, namely that of temperature and entropy.

A reaction is spontaneous if $ \Delta G < 0 $ such that:

If $ \Delta H < 0$ and (-) and $ \Delta S > 0$ (+), the reaction is spontaneous at any temperature.

If $ \Delta H > 0$ and (+) and $ \Delta S < 0$ (-), the reaction is NOT spontaneous at any temperature.

If $ \Delta H < 0$ and (-) and $ \Delta S < 0$ (-), the reaction is spontaneous at low enough temperatures.

If $ \Delta H > 0$ and (+) and $ \Delta S > 0$ (+), the reaction is spontaneous at high enough temperatures.

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  • $\begingroup$ So I can't plot this delta_G_1 = delta_H_1 - T * delta_S_1 as a function of T because the experimental tabulated values are true just for the temperature of 25 °C right? I've added also the corrected version of the calculations, is it correct? $\endgroup$ – Aurelius Mar 2 '14 at 0:00
  • $\begingroup$ @FormlessCloud Exactly, because those values indexed inside the appendix are derived from experiments. If temperature were to change during those heats of formation, we would have to run an experiment to determine them. However, we can run reactions at different temperature. Those look right. I would sure that they are for the correct state. A good example is water in liquid and gas phase. $\endgroup$ – Jun-Goo Kwak Mar 2 '14 at 0:03
  • $\begingroup$ But if the experimental tabulated values are going to change at different temperatures than 25 °C how we can predict the spontaneity of the reacton in such a way? I mean this way, by temperature: (-) and (+), the reaction is spontaneous at any temperature. (+) (-), the reaction is NOT spontaneous at any temperature. (-) (-)the reaction is spontaneous at low enough temperatures. (+) (+) the reaction is spontaneous at high enough temperatures. $\endgroup$ – Aurelius Mar 3 '14 at 18:02
  • $\begingroup$ @FormlessCloud Hi, I was referring to the calculation of the standard thermodynamic data. Entropy and enthalpy are experimentally determined at $25^\circ C$. If we run the entire reaction, we would just change the T term in the equation above. Sorry for the confusion. $\endgroup$ – Jun-Goo Kwak Mar 3 '14 at 18:06
  • $\begingroup$ Hi, don't worry : ) However ok, but if are not allowed to change the temperature because the other constants of the equation are no longer valid at different temperature than 25 °C, how this can be possible? From what I understood then the $\Delta G$ can just tell us that the raction is spontaneous or not at the temperature of 25 °C. If we say that since $\Delta H$ and $\Delta S$ are both positive (+) the reaction is spontaneous at high enough temperatures maybe at that "high enough temperatures" $\Delta H$ and $\Delta S$ can change their signs, right? $\endgroup$ – Aurelius Mar 3 '14 at 18:17

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