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When $\pu{20.00 mL}$ of $\pu{1.00 M}$ $\ce{AgNO3}$ solution is added to $\pu{20.00 mL}$ of $\pu{1.00 M}$ of $\ce{NaI}$ at $\pu{25 ^\circ C}$ in a calorimeter, a white precipitate of $\ce{AgI}$ is formed. The temperature of the aqueous mixture increases to $\pu{40 ^\circ C}$.

I am trying to calculate the $\Delta H$ for the reaction per mole of $\ce{AgI}$. The specific heat of the aqueous mixture is $\pu{4.184 J//g K}$, the density of the mixture $\rho = \pu{1.00 g//ml}$. And I assume that the calorimeter absorbs a negligible amount of heat.

How am I suppose to go about finding the $\Delta H$ per mole of $\ce{AgI}$? I don't really know where to start here as I am unclear about the whole problem here. In order to use the formula $$Q = m C_p \Delta T$$ I need to find the mass. But which mass do I take? $\ce{AgI}$, or $\ce{NaNO3}$, or is it the mass of the entire product of the reaction?

This is the balanced equation $$\ce{AgNO3 + NaI -> AgI + NaNO3}$$

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    – Martin - マーチン ♦
    Dec 12, 2017 at 16:46

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When the reaction takes place and releases its heat, everything in the calorimeter heats up: the water, the $\ce{AgI}$ and the $\ce{NaNO3}$. But of these the water is by far the most massive and for that reason, as a reasonable approximation, we consider that mass that absorbs the heat to be the water, i.e. $40.00\ \mathrm{g}$.

Use this to calculate $Q$.

Then determine how many moles of $\ce{AgI}$ was formed, by stoichiometry. The quotient of these two numbers gives you the reaction enthalpy per mole of $\ce{AgI}$

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  • $\begingroup$ So, for Q I got 2510.4J (using mass as 40g as we assume no changes in volume. By mole ratio, I got 0.02 Moles AgI . So the question wants per mole of AgI . So am I right to say that 0.02 moles of AgI makes 2510.4 J ? And now I just need to find 1 mole of AgI make much heat in joules ? $\endgroup$
    – user175089
    Dec 13, 2017 at 1:44
  • $\begingroup$ Yes. Note that "we assume no changes in volume" is irrelevant: this is about mass, not volume. $\endgroup$
    – Gert
    Dec 13, 2017 at 13:33

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