Exothermic reaction in adiabatic reactor

[MexChemE]

Homework Statement

One mole of gas A, two moles of gas B, and one mole of inert gas I are fed into an adiabatic reactor of variable volume and constant pressure at 25 °C. At this temperature, the reaction yielding liquid R proceeds normally as:
\textrm{A} (g) + \textrm{B} (g) \rightarrow \textrm{R} (l)
However, the reaction is exothermic, and all products are heated up to 325 °C. R boils at 125 °C. Find ΔH_Reaction for the given reaction.
C_P,A(g) = 30 J mol^-1 K^-1
C_P,B(g) = 40 J mol^-1 K^-1
C_P,I(g) = 30 J mol^-1 K^-1
C_P,R(l) = 60 J mol^-1 K^-1
C_P,R(g) = 80 J mol^-1 K^-1
ΔH_R,lg = 10,000 J mol^-1

Homework Equations

Q_{\textrm{net}}=0
Q=nC_P \Delta T

The Attempt at a Solution

Since the reactor is adiabatic, all the products are heated with the energy released by the reaction:
\Delta H_{\textrm{Reaction}} + Q_{\textrm{B}} + Q_{\textrm{I}} + Q_{\textrm{R,l}} + \Delta H_{\textrm{R,lg}} + Q_{\textrm{R,g}} = 0
Where Q represents the energy required to heat each substance from 25 to 325 °C, except for R which first heats from 25 to 125 °C, boils, and then heats from 125 to 325 °C.

After calculating and adding all sensible and latent heats I got 53,000 J, as the reactor is adiabatic, the heat released by the reaction was absorbed and distributed among the products, which caused the temperature to rise. Therefore:
\Delta H_{\textrm{Reaction}} = -53 \ \textrm{kJ}
Is there a more sophisticated way to do this?

 

[Chestermiller]

Is it correct to say that you are trying to determine the heat of reaction at 25 C?

Chet

 

[MexChemE]

I guess not. It's actually the heat of reaction happening from 25 °C to 325 °C, right? I guess we could also use Kirchhoff's law because we have the specific heats of reactants and products, but we're missing a heat of reaction to use as our reference point.

 

[Chestermiller]

I guess not. It's actually the heat of reaction happening from 25 °C to 325 °C, right? I guess we could also use Kirchhoff's law because we have the specific heats of reactants and products, but we're missing a heat of reaction to use as our reference point.

Conventionally, heat of reaction is a quantity defined at a specified temperature, with reactants and products at the specified temperature. My understanding is that you are trying to take the results of an experiment in an adiabatic reactor, and back out from these results the heat of reaction at a certain temperature (probably 25C).

Chet

 

[Chestermiller]

I would do this differently. I would work starting with heats of formation at 25C. It is less prone to errors, and let's the mathematics do all the work for you. Since the pressure is held constant, the change in enthalpy is equal to the heat added. I would get the enthalpy of the mixture in the initial state, and the enthlypy of the mixture in the final state. The differences between the heats of formation at 25 C corresponding to the heat of reaction automatically falls out of this.

Chet

 

[MexChemE]

I would do this differently. I would work starting with heats of formation at 25C. It is less prone to errors, and let's the mathematics do all the work for you. Since the pressure is held constant, the change in enthalpy is equal to the heat added. I would get the enthalpy of the mixture in the initial state, and the enthlypy of the mixture in the final state. The differences between the heats of formation at 25 C corresponding to the heat of reaction automatically falls out of this.

Chet

Yes, that would seem to be the easiest, and most exact way, although this specific problem didn't provide the heat of formation of products and reactants, nor an enthalpy of reaction at a reference state.

 

[Chestermiller]

Yes, that would seem to be the easiest, and most exact way, although this specific problem didn't provide the heat of formation of products and reactants, nor an enthalpy of reaction at a reference state.

The heats of formation appear algebraicly in the final equation in the exact right combination to allow determination of the heat of reaction. Try it and see.