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Right now I'm doing a calorimetry lab and for the pre-lab we have been asked to determine the temperature change (in Celsius) of ammonium chloride in water.

The question asks for the expected temperature change ($\Delta T$) of $\pu{8.5 g}$ of $\ce{NH4Cl}$ in $\pu{100 mL}$ (or $\mathrm{g}$) of water, with the molar enthalpy ($\Delta H_\mathrm{sol}$) of the solution being $\pu{0.277 kJ/g}$. The conversion of $\mathrm{kJ/g}$ is throwing me off and I cannot figure out how to solve for $\Delta T$ with the given info.

I know that there are $0.165048...$ moles of solution, which gives me everything I need to solve. We were given the equation

$$n \Delta H_\mathrm{sol} = mC \Delta T,$$

where $m$ - mass of water and $C$ - specific heat capacity of water. I am assuming it needs to be rearranged to

$$\Delta T = \frac{mC}{n\Delta H_\mathrm{sol}}$$

Any help is greatly appreciated and I can explain in further detail if necessary. Sorry for the Celsius, we apparently don't use Kelvin in our calculations.

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    $\begingroup$ There is no such thing as mole of solution. $\endgroup$ – Ivan Neretin Dec 6 '16 at 5:32
  • $\begingroup$ @IvanNeretin Sure there is. If I have a mixture of chemicals that sum to 6.022 x 10^23 molecules, then I have one mole of solution. $\endgroup$ – Bob May 24 '17 at 2:28
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The main problem here is a simple mistake in the algebra. You rearranged:

$\pu{n\timesΔH_{sol} = m\times C\timesΔT}$

to

$\pu{ΔT = \frac{m\times C}{n\timesΔH_{sol}}}$

rather than

$\pu{ΔT = \frac{n\timesΔH_{sol}}{m \times C}}$

Additionally, the molar enthalpy of solvation is given in the problem as units of kJ/g, not kJ/mol. Presumably, this is an error in the given problem. According to Parker, V . B ., Thermal Properties of Uni-Univalent Electrolytes, Natl . Stand . Ref . Data Series — Natl . Bur. Stand .(U.S.), No .2, 1965, the molar enthalpy of solution for $\ce{NH4Cl}$ is $\pu{14.78 kJ/mol}$.

There was also a slight miscalculation in the moles of solute. Where you calculated $\pu{0.165 moles}$ of $\ce{NH4Cl}$, you should have gotten:

$\mathrm{8.5 g / 53.49\frac{g}{mol}=0.159~mol}$

These errors corrected, plugging in the values to solve for $\Delta \text{T}$ is trivial and gives:

$\pu{ΔT = \frac{\pu{0.159 mol}\times \pu{14.78 kJ mol-1}}{\pu{100g} \times \pu{4.186J~g-1~K-1}} = 5.6K}$

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I agree with Airhuff (almost) completely.

It doesn't really matter if you use kJ/mol, or kJ/g. As long as your units can cancel out.

$$\frac{14.78kJ/mol}{53.491g/mol} = 0.277kJ/g$$

But I mostly wanted to point out, the reason why it is okay to use Celsius for this calculation - because you have "ΔT" in your expression.

Say you have something at 30°C, and it changes to 24°C.

ΔT = 24°C - 30°C = -6°C

In Kelvin, the temperatures are 303K or 297K.

ΔT = 297K - 303K = -6K

The way I would do it: $$ΔT =8.5g NH4Cl * \frac{0.277kJ}{1gNH4Cl} * \frac{1}{100g H2O} *\frac{1gH2O * °C}{0.004184kJ} = 5.627°C$$

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