Question

What is the change in internal energy of reaction when `"0.721 g"` of titanium is placed into a bomb calorimeter and allowed to combust, if the temperature rises from `25^@ "C"` to `53.8^@ "C"`? `C_"cal" = "9.84 kJ/K"`.

`A)` `"283.4 kJ/mol"`
`B)` `-"283.4 kJ/mol"`
`C)` `1.89 xx 10^4 "kJ/mol"`
`D)` `-1.89 xx 10^4 "kJ/mol"`

Answer 1

I'm getting `D`.

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A bomb calorimeter by definition is a constant-volume calorimeter. These are rigid and closed, generally with a large water bath to insulate the system.

This is in such a way that heat generated or absorbed due to the reaction is equal to the heat absorbed or generated (respectively) by the calorimeter (conservation of energy).

![https://www.learner.org/]()

That means you must recall that `DeltaE = q_V`, where `q_V` is the heat flow `q` at constant volume, since `DeltaE = q + w = q - cancel(PDeltaV)^(0)`.

The following data was given:

• `m_"Ti" = "0.721 g"`

• `DeltaT = T_2 - T_1 = 53.8^@ "C" - 25^@ "C" = 28.8^@ "C"`

• `C_"cal" = "9.84 kJ/K"`, or `"9.84 kJ/"^@ "C"`
  (note: `DeltaT` in `""^@ "C"` is the same as `DeltaT` in `"K"`, but the individual temperature values are NOT.)

Since you were given the heat capacity of the calorimeter, you can calculate `q_V` for the calorimeter itself:

`color(green)(q_(V,"cal")) = m_"cal" c_"cal"DeltaT`

`= C_"cal"DeltaT`

`= ("9.84 kJ/"^@ "C")(28.8^@ "C")`

`=` `color(green)("283.4 kJ")`

where `c` is the specific heat capacity in `"kJ/mol"^@ "C"` and `C` is the heat capacity in `"kJ/"^@ "C"`. Do not confuse them---they are not the same!

So, we can set this equal to `DeltaE_"cal"` as we stated earlier:

`DeltaE_"cal" = q_(V,"cal") = "283.4 kJ"`

Finally, in `"kJ/mol Ti"`, we must calculate the `"mol"`s of `"Ti"` combusted:

`n_"Ti" = m_"Ti"/(M_(m,"Ti"))`

`= "0.721 g" xx "1 mol Ti"/"47.867 g" = "0.0151 mols Ti"`

In the end, calculating the change in internal energy `DeltaE` (some might call it `DeltaU` to not confuse it with the total energy) for the reaction, which is also the heat of reaction in this case, would ask you to realize that:

`q_(V,"rxn") = -q_(V,"cal") = -"283.4 kJ"`

when energy is conserved. Therefore, for the reaction, we have that:

`color(blue)(DeltaE_"rxn") = q_"rxn"/(n_"Ti")`

`= -"283.4 kJ"/("0.0151 mols Ti")`

`= -"18814.3 kJ/mol"`

`= color(blue)(-1.89 xx 10^4)` `color(blue)("kJ/mol")`