# Why is this calculation of the enthalpy of combustion wrong?

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1. Nov 14, 2016

### Buffu

1. The problem statement, all variables and given/known data
A 1.250 g sample of $C_8H_18$ is burnt with excess oxygen in a bomb calorimeter. Change in temperature is from 294.05 K to 300.78 K. If heat capacity of the calorimeter is 8.93 kJ/K. Find the heat transferred to the calorimeter. Also calculate the enthalpy of combustion of the sample of octane.

2. Relevant equations
$q_v = \triangle U = C \times \triangle T$
$\triangle H = \triangle U + \triangle n_{(g)}RT$
$C_8H_{18} (s) + {25\over 2}O_2 \longrightarrow 8CO_2(g) + 9H_2O(l)$

3. The attempt at a solution
I use the first equation to find $q_{transferred}= (300.78 - 294.05) \times 8.93 J = 60.1 kJ$
Since combustion is exothermic, Thus $q_{transferred} = 60.1 kJ$.

Since this reaction is carried the volume is constant but pressure isn't.

Since $-q_{transferred} = \triangle U = -60.1 kJ$
Now enthalpy of combustion is $\triangle H = \triangle U + \triangle n_{(g)}RT$

$\triangle n_{(g)} = (8 - 12.5) \times {1.25 \over 114}$

Since $-q_{transferred} = \triangle U = -60.1 J$ for exothermic reactions

$\therefore \triangle H = -60.1 + 8.314 \times 294.05 \times (8 - 12.5) \times {1.25 \over 114} \times 10^{-3}$
Which i got from calculator as $-60.221 kJ$.

Now my second answer is incorrect.

In the solution $\triangle U$ is taken positive and $\triangle_{combustion} H = {\triangle U \over \text{number of moles}}$

I don't understand the reasoning behind taking $\triangle U$ positive as combustion is exothermic and thus heat is evolved, which by convention means that we need to get -ve sign in the answer.
For $\triangle H$ calculation by the author, I don't even have a clue how they did that.

Last edited: Nov 14, 2016
2. Nov 15, 2016

### Staff: Mentor

Did you say that you got the heat transferred to the calorimeter correct?

3. Nov 15, 2016

### Buffu

Yes as per the book. I think it should be -ve but, I am not sure.

4. Nov 15, 2016

### Staff: Mentor

Well, the heat transferred to the calorimeter is positive. The heat transferred to the reaction mixture is negative.

You correctly determined the change in enthalpy of the reaction mixture in the experiment. But, the heat of combustion, is defined as the change in enthalpy per mole. So you need to divide by the number of moles to get the heat of combustion. And, your interpretations regarding sign of the heat of combustion as minus the enthalpy change are correct.

5. Nov 15, 2016

### Buffu

So all I need to do is $-60.221\over ({1.25 \over 114})$ to get the correct answer.
Thank you very much.