Reaction enthalpy [closed]








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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?




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
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> 


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. 


$$q\_\mathrm{reaction} + q\_\mathrm{calorimeter} + q\_\mathrm{water} = 0$$



> 
> $q\_\mathrm{calorimeter}$
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You calculate the heat absorbed by the calorimeter correctly. 


$$q\_\mathrm{calorimeter} = C\_\mathrm{calorimeter} \Delta T = \mathrm{ \left( 1251\ \dfrac{J}{K} \right) \left(3.5\ K\right) = 4379\ J}$$



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> $q\_\mathrm{water}$
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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. 


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



> 
> $q\_\mathrm{reaction}$
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> 
> 


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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