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In an experiment, we measured the temperature difference in a reaction of $2.5\ \mathrm g$ of $\ce{Na2SO3}$ with $25\ \mathrm{ml}$ of $10{-}13\ \%$ $\ce{NaClO}$ in a calorimeter with $300\ \mathrm{ml}$ of water.

Now I am to calculate the enthalpy of the reaction, via the thermodynamic data provided as well as via through the experimental results.

Using the provided data:

$$\Delta H_\mathrm f[\ce{SO4^2-(aq)}] = -909\ \mathrm{kJ/mol}$$

$$\Delta H_\mathrm f[\ce{SO3^2-(aq)}] = -636\ \mathrm{kJ/mol}$$

$$\Delta H_\mathrm f[\ce{ClO-(aq)}] = -107\ \mathrm{kJ/mol}$$

$$\Delta H_\mathrm f[\ce{Cl-(aq)}] = -167\ \mathrm{kJ/mol}$$

$$\Delta H/\mathrm{mol} = \Delta H(\text{products}) - \Delta H(\text{reactants}) = 333\ \mathrm{kJ/mol}$$

Number of moles of the products $0.0397150\ \mathrm{mol}$

$$\Delta H(\text{reaction}) = \Delta H/\mathrm{mol} \times 0.0397150\ \mathrm{mol}= 13.22510096\ \mathrm{kJ}$$

Using the measured temperature difference:

$$q(\text{reaction}) = C(\mathrm{cal})\cdot\Delta T = 1251.4344\ \mathrm{J/K}\times3.5\ \mathrm K = -3754.3032\ \mathrm{J}$$

$$\Delta H(\text{reaction}) = q(\text{reaction})/\ \mathrm{mol}\ (\ce{Na2SO3})= -3754.3032\ \mathrm{J}/0.01986\ \mathrm{mol} = -189062.1432\ \mathrm{J/mol}$$

Which value is the correct one?

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closed as unclear what you're asking by Zhe, Todd Minehardt, Jan, ron, Curt F. Jan 3 '17 at 22:56

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  • $\begingroup$ What's the question? And what about the ~25ml water from the hypochlorite solution, and the calorimeter heat capacity? $\endgroup$ – aventurin Jan 3 '17 at 15:29
  • $\begingroup$ The results are completely different. The calorimeter heat capacity is C(cal) = 1251,4344 $\endgroup$ – Anna Sarah Jan 3 '17 at 15:46
  • $\begingroup$ 1256 kJ/K is the heat capacity of 300ml water. This does neither include the ~25ml water from the hypochlorite solution nor the heat capacity of the calorimeter itself. So it's not surprising that the result differs significantly from the expected value. $\endgroup$ – aventurin Jan 3 '17 at 15:59
  • $\begingroup$ So it seems that I have problems with understanding the term "heat capacity" - I thought the amount of water in the calorimeter was all I needed and I wasn't aware I had to account for the Hypochlorite as well. Thanks for the hint, I will have another look. $\endgroup$ – Anna Sarah Jan 3 '17 at 16:05
  • $\begingroup$ Units added - I have experimentes with different options, but would think that this could only account for a comma shift, but not entirely different values. $\endgroup$ – Anna Sarah Jan 3 '17 at 16:46
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I get 460 kJ/mol using your data, after making an assumption. Not too far off from 333 kJ/mol. Here's how we get there.

First law of thermodynamics

If we pretend that your calorimeter is an isolated system (and the calorimeter constant helps us handle the flaw in this assumption) and that no work is done, then the sum of all heat transfers must be zero. You started here, but it looks like you forgot the heat absorbed by the water.

$$q_\mathrm{reaction} + q_\mathrm{calorimeter} + q_\mathrm{water} = 0$$

$q_\mathrm{calorimeter}$

You calculate the heat absorbed by the calorimeter correctly.

$$q_\mathrm{calorimeter} = C_\mathrm{calorimeter} \Delta T = \mathrm{ \left( 1251\ \dfrac{J}{K} \right) \left(3.5\ K\right) = 4379\ J}$$

$q_\mathrm{water}$

Water requires $\mathrm{4.184\ J/g}$ to increase its temperature by $\mathrm{1\ K}$. You have approximately $\mathrm{325\ g}$ of water from the two solutions.

$$\mathrm{q_{water} = ms\Delta T = \left( 325\ g \right)\left(4.184\ \dfrac{J}{gK}\right)\left(3.5\ K\right)= 4759\ J}$$

$q_\mathrm{reaction}$

The reaction must have produced $\mathrm{-9138\ J}$. When you divide by the number of moles you give is $\mathrm{-460.3\ kJ}$. Huh. Wrong sign. Are you certain about your $\Delta T$? Are you sure you calculated it by $\Delta T = T_\mathrm{final}-T_\mathrm{initial}$ and not as the unsigned difference between the larger and smaller values? If I assume your temperature change was negative, I get the correct sign.

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  • $\begingroup$ I am certain about the temperature, which was rising, giving a positive ΔT. I believe that the calorimeter is not to be considered after all, just the water, but I am not sure. $\endgroup$ – Anna Sarah Jan 3 '17 at 22:32

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