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If I've found the change in enthalpy of an explosion reaction, how would I go about finding the final temperature and pressure in a constant volume using the enthalpy of reaction, number of moles of reactants, and a given constant volume heat capacity if all the products are ideal gases?

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2 Answers 2

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$\Delta H = \Delta(U + PV) = \Delta U + V\Delta P$, given a constant volume.

We also have the relations $\Delta U = C_{v} \Delta T$ and $PV = nRT \implies \Delta P = \frac{nR}{V} \Delta T$.

Substituting gives $\Delta H = C_{v} \Delta T + nR\Delta T = (C_{v} + nR) \Delta T \implies \Delta T = \frac{\Delta H}{C_{v} + nR}$. You can then substitute this result back into the relation between $\Delta T$ and $\Delta P$ to find $\Delta P$.

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For an ideal gas reaction, the standard heat of reaction $\Delta H^0$ represents the amount of heat that must be added to the reaction mixture at constant pressure of one bar in order for the final temperature to equal the initial temperature of 25 C. For this same change, the internal energy change $\Delta U^0$ is given by $$\Delta U^0=\Delta H^0-\Delta PV=\Delta H^0-(\Delta n)RT^0$$ where $\Delta n$ is the change in the number of moles in the reaction. Since, for an ideal gas mixture, the internal energy is independent of pressure and volume, the internal energy of the final state would not change if we returned the products to the initial volume. So $\Delta U^0$ also represents the change in internal energy in going from reactants to products at constant volume and temperature $T^0$. If the reaction were carried out adiabatically and at constant volume, no work would be done and no heat transfer would occur. So, for this situation, the change in internal energy would be zero. So, we would have:

$$\Delta U=\Delta U^0+n_pC_v(T_f-T^0)=0$$ where $n_v$ is the number of moles of products and $C_v$ is the heat capacity at constant volume of the product mixture. So, combining the previous equations, we would have: $$T_f=T^0-\frac{(\Delta H^0-(\Delta n)RT^0)}{n_pC_v}$$

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