How does enthelpy change as the reaction proceeds?

[DrDu]

Thermodynamically, it makes not much sense to distinguish between forward and backward reactions which are rather concepts from kinetics. All you see is a net reaction, i.e. a change of the concentration of the product with time. If this net rate is $r_i=dn_i/dt$, then $d\xi=r_i/\nu_idt$ and the heat produced per time step is $dH=\Delta H d\xi$. Of course if temperature is changing, too, then you have to take this into account as Chet did in his last post.

 

[sgstudent]

Thermodynamically, it makes not much sense to distinguish between forward and backward reactions which are rather concepts from kinetics. All you see is a net reaction, i.e. a change of the concentration of the product with time. If this net rate is $r_i=dn_i/dt$, then $d\xi=r_i/\nu_idt$ and the heat produced per time step is $dH=\Delta H d\xi$. Of course if temperature is changing, too, then you have to take this into account as Chet did in his last post.

Yes that makes sense. But isn't the rate of reaction not constant because of the mixture reaction equilibrium causing the concentrations of the reactants and products to change? And so dH would change too?

Something like the concentration of products increase over the extent of reaction and so the backwards reaction being more prominent causes the overall dH to decrease? Sorry if I'm bringing this into a loop I just can't seem to get why concentration isn't a factor in dH.

 

[Chestermiller]

And as Chestermiller and you have agreed ΔH is affected by both the forward and backwards reaction. So as the reaction goes on wouldn't the backwards reaction decrease the ΔH as well?

We seem to be talking about two different concepts here. One of these is the enthalpy variation of the reaction mixture, and the other is the molar "Heat of Reaction" of an individual reaction. The enthalpy of a reaction mixture will change as the concentrations of the species in the reactor changes, but the molar Heat of Reaction for each individual reaction of the mixture will not change as the concentration of the species in the reactor changes (if the mixture is an ideal gas mixture or an ideal fluid mixture, each of which exhibits a heat of mixing of zero). In my previous post, I expressed the molar Heat of each Reaction in terms of the partial molar enthalpies of the reactants and products comprising that particular reaction. For an ideal solution, the partial molar enthalpies of the reactants and products are equal to the molar enthalpies of the pure reactants and products at the same temperature and total pressure as the mixture. So the molar heat of an individual reaction is the same in the reaction mixture as it is for going from stoichiometric molar quantities of the pure reactants and ending up with corresponding stoichiometric molar quantities of the pure products.

If the molar Heat of Reaction for an individual reaction changed significantly with concentration even for ideal gas and liquid solutions, then there would be no point in preparing tables of heats of reaction and heats of formation, because they couldn't be used for anything. Their very existence tell us that, at least for ideal solutions, the molar Heats of Reaction do not change with concentration.

Chet

 

[Chestermiller]

Something like the concentration of products increase over the extent of reaction and so the backwards reaction being more prominent causes the overall dH to decrease? Sorry if I'm bringing this into a loop I just can't seem to get why concentration isn't a factor in dH.

From my post #28, for a single reversible reaction occurring in a reactor held at constant temperature, the equation for the variation of the mixture enthalpy H with time is given by:

$$\frac{dH}{dt}=V(r_fΔH_R-r_rΔH_R)=V(r_f-r_r)ΔH_R$$

where V is the reactor volume, r_f is the rate of the forward reaction, r_r is the rate of the reverse reaction and ΔH_R is the molar Heat of Reaction (which is independent of concentration for an ideal solution). So,while the enthalpy of the reaction mixture is changing with time, the molar Heat of Reaction is constant.

Chet

 

[DrDu]

From my post #28, for a single reversible reaction occurring in a reactor held at constant temperature, the equation for the variation of the mixture enthalpy H with time is given by:

$$\frac{dH}{dt}=V(r_fΔH_R-r_rΔH_R)=V(r_f-r_r)ΔH_R$$

where V is the reactor volume, r_f is the rate of the forward reaction, r_r is the rate of the reverse reaction and ΔH_R is the molar Heat of Reaction (which is independent of concentration for an ideal solution). So,while the enthalpy of the reaction mixture is changing with time, the molar Heat of Reaction is constant.

Chet

... and $r=r_f-r_r$ is the net rate of the reaction.

 

[sgstudent]

From my post #28, for a single reversible reaction occurring in a reactor held at constant temperature, the equation for the variation of the mixture enthalpy H with time is given by:

$$\frac{dH}{dt}=V(r_fΔH_R-r_rΔH_R)=V(r_f-r_r)ΔH_R$$

where V is the reactor volume, r_f is the rate of the forward reaction, r_r is the rate of the reverse reaction and ΔH_R is the molar Heat of Reaction (which is independent of concentration for an ideal solution). So,while the enthalpy of the reaction mixture is changing with time, the molar Heat of Reaction is constant.

Chet

But wouldn't the rate of the forward and backwards reaction also change with time?

... and $r=r_f-r_r$ is the net rate of the reaction.

It seems like you and Chestermiller have a different concepts on ΔH. So I'm a bit confused. Is ΔH fixed or does it change during the reaction?

 

[DrDu]

It seems like you and Chestermiller have a different concepts on ΔH. So I'm a bit confused. Is ΔH fixed or does it change during the reaction?

No, our concepts are quite coincident.
$\Delta H$ changes with reaction are mostly negligible for all practical purposes.

 

[Chestermiller]

But wouldn't the rate of the forward and backwards reaction also change with time?

Sure. But, for an ideal gas mixture or an ideal liquid solution, ΔH_R doesn't change.

So what if the forward and backward reaction rates change? That's a separate factor from the ΔH_R.

Chet