0
$\begingroup$

In a constant volume bomb calorimeter, we measure the change in temperature of the calorimeter and find out it's change in internal energy. We then convert it to change in enthalpy.

To determine $\Delta H_c$ we write the balanced chemical equation to find $\Delta n_g$ and using the value of $\Delta U$ obtained from calorimeter, we apply $$\Delta H = \Delta U + \Delta n_g RT$$

The above equation is only valid for constant pressure and temperature. How can we apply this equation on calorimeter, where the temperature change is observed?

$\endgroup$
4
  • $\begingroup$ What do you mean by how can we apply the equation? if you have the numbers then all you have to is plug them in $\endgroup$
    – Buraian
    Oct 6 '20 at 12:07
  • $\begingroup$ The main requirement of the equation is that temperature should be constant throughout the reaction. That's how it is derived. But temperature changes in a calorimeter. So this equation should be invalid here. $\endgroup$
    – Eyy boss
    Oct 6 '20 at 12:16
  • $\begingroup$ I'm not sure from where you've studied it but the formula is valid in a constant volume bomb calorimeter. There is no particular conditions on pressure and temperature.. where did you get that from? $\endgroup$
    – Buraian
    Oct 6 '20 at 12:33
  • $\begingroup$ From the basic definiton, $\Delta H = \Delta U + \Delta (pV)$. If we consider ideal behavior, then $\Delta (pV) = \Delta (nRT)$. This will become $\Delta n_g RT$ only when $T$ is constant. $\endgroup$
    – Eyy boss
    Oct 6 '20 at 12:55
1
$\begingroup$

In a constant volume bomb-calorimeter, we can make chemical reactions take place. If we let chemical reactions freely take place in it, then the total internal energy change is zero. This is due to the fact there is no expansion work since the system is a constant volume by definition and the heat transferred is zero since the bomb calorimeter is well insulated. The heat released in a reaction is automatically absorbed by the bomb calorimeter device. Hence the total internal energy change is zero.

Now, consider another path of the reaction. We start with reactants and turn them into products under constant volume and constant temperature conditions (*) and then these products we raise the temperature as to reach the final temperature we got in the previous path.

The energy associated with moving from reactants to products isothermally is $\Delta U_{rcxtn}$ and the energy change for raising up the products to the temperature of first path is $ \Delta U_{heat}$

Now, we can write:

$$ \Delta U_{heat} = C_{v}^{cal} \Delta T$$

And,

$$ \Delta U_{rcxtn} + \Delta U_{heat} = 0$$

Or,

$$ \Delta U_{rcxtn} = - C_{v}^{cal} \Delta T$$

Now, consider the enthalpy change of reaction:

$$ \Delta H_{rcxtn} = \Delta U_{rcxtn} + \Delta (PV)$$

Now, if we were to assume our reactants as ideal gases a, then $PV=nRT$ and also considering constant temperature, we get:

$$ \Delta (PV) = \Delta (nRT) = RT \Delta n$$

Using the above result and expression for internal energy, we finally arrive at the following equation:

$$ \Delta H_{rcxtn} = - C_{v}^{cal} \Delta T + RT \Delta n$$


Refernces:

Around 42:49 of this video

Page-2 of these notes

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.