Enthalpy of Solution of Copper Sulphate

Transcript

0:00Introduction

Hello everyone! Here is a small lab puzzle. Stir anhydrous copper sulphate, the salt with no water in its crystals, into water, and the water gets warmer. Stir in the pentahydrate, the same salt carrying five water molecules, and the water gets slightly colder! Same copper sulphate, opposite results. So why? And how do we measure these heats with just a vessel and a thermometer? This experiment is part of the practical chemistry in the JEE Main syllabus. So, let's begin.

0:34The calorimeter and the heat equation

So, here is the apparatus from the course figure: an insulated vessel with a lid, a thermometer reading to zero point one degree Celsius, and a stirrer. When the dissolving salt gives out heat, who takes that heat? Yes, two things take it: the solution, and the vessel itself. So the heat, small q, equals m times c plus capital C cal, all times delta T, okay? The little triangle is the Greek capital letter delta, so delta T is the change in temperature. Small m is the mass of the solution, and small c is its specific heat, taken as that of water: four point one eight joules per gram per kelvin. The raised minus ones mean per gram and per kelvin. Clear? And capital C with the small subscript cal, short for calorimeter, is the heat capacity of the vessel, also called the calorimeter constant. With me?

1:27First, find the calorimeter constant

But how do we know capital C cal? We measure it first: mix known masses of warm and cold water in the calorimeter. The cold water gains less heat than the warm water loses. So where did the rest go? Yes, exactly, it went into the vessel! So, heat lost by the warm water equals heat gained by the cold water, plus capital C cal times delta T, and that gives us capital C cal. Right?

1:54Enthalpy of solution and its sign

Now, what are we measuring? The enthalpy of solution, delta H with a small subscript sol. It is the enthalpy change when one mole of a substance dissolves in a large excess of solvent. Okay? So we divide the heat by small n, the moles dissolved. Delta sol H equals minus or plus, q over n; that sign has the minus written above the plus. Got it? Now, suppose the temperature rises. Many students write a plus sign, because the temperature went up. So which is it, plus or minus? It is minus, and the plus was the trap. The temperature rose because the salt gave out heat, so the change is exothermic, and delta sol H is negative. If the temperature falls, the sign is plus. Clear?

2:40Getting the true temperature change

Two practical points now. The salt is finely powdered, added all at once, and stirred. Why all this care? So that it dissolves quickly and evenly, because a slow process loses more heat to the surroundings and gives a smaller temperature change. Okay? Now the graph: temperature going up, time going across. After mixing, the temperature climbs to a peak, then falls slowly along the cooling line. So why not just take the highest reading? Because the mixture loses heat while the salt is still dissolving, so the highest reading is below the true value. So we produce the cooling line back to the time of mixing, the orange dashed line, and the double arrow marked delta T gives the true change. With me?

3:28Copper sulphate: why the two salts differ

Now, back to our puzzle. Anhydrous copper sulphate, C u S O four, dissolves with evolution of heat, about minus sixty six kilojoules per mole. Right? The pentahydrate, with a dot and five H two O for five water molecules in the crystal, dissolves with a slight absorption of heat, about plus twelve kilojoules per mole. So why the difference? For the anhydrous salt, the hydration of the ions releases more energy than is needed to break up the lattice, so heat comes out. Okay? In the pentahydrate, the ions are already largely hydrated, so little hydration energy is left to offset the lattice energy. That solves our puzzle! Got it?

4:09Hess's law: the enthalpy of hydration

Now, the arrow shows anhydrous copper sulphate becoming the pentahydrate, and the heat of this change is the enthalpy of hydration. So why can we not measure it directly? Because the solid takes up exactly five molecules of water only slowly and unevenly, so the heat cannot be measured. Right? So we use Hess's law, and combine two heats of solution that are easy to measure. Delta H with the subscript hyd, for hydration, equals delta sol H of the anhydrous salt, minus delta sol H of the hydrate. Clear?

4:44Example 1: calorimeter constant from warm and cold water

Alright, example one. Fifty grams of water at forty five point zero degrees Celsius is added to fifty grams of water at twenty five point zero degrees Celsius in a calorimeter. The final temperature is thirty four point zero degrees Celsius. Find the calorimeter constant. Pause the video and try it first. Done? Okay, let's check. The warm water falls by eleven degrees, which is eleven kelvin. So it loses fifty times four point one eight times eleven point zero, which gives two thousand two hundred and ninety nine joules. Right? The cold water rises by nine degrees, so it gains fifty times four point one eight times nine point zero, which gives one thousand eight hundred and eighty one joules. The rest went into the calorimeter. But did it warm by eleven kelvin, or by nine? It warms by nine kelvin, and many students slip here: it went from twenty five to thirty four degrees, with the cold water. So it took two thousand two hundred and ninety nine minus one thousand eight hundred and eighty one, which is four hundred and eighteen joules, for nine point zero kelvin. Okay? So capital C cal is four hundred and eighteen over nine point zero, which gives forty six point four joules per kelvin. Keep it handy for the next example. With me?

6:02Example 2: enthalpy of solution of anhydrous CuSO4

Now a harder one, with the same calorimeter. Four point zero grams of anhydrous copper sulphate is dissolved in one hundred grams of water. Its molar mass, capital M, is one hundred and fifty nine point five. The temperature rises by three point five kelvin. Find the enthalpy of solution, neglecting the heat capacity of the salt. Pause the video and try it first. Done? Okay, let's check. The water takes one hundred times four point one eight, which is four hundred and eighteen joules per kelvin. Add the calorimeter's forty six point four, and multiply by three point five. That gives one thousand six hundred and twenty five joules. Clear? The moles are four point zero over one hundred and fifty nine point five, which gives zero point zero two five zero eight. The temperature went up. So is the answer plus or minus? It is minus, since heat was evolved. So delta sol H is minus one thousand six hundred and twenty five joules over zero point zero two five zero eight moles, which gives minus sixty four point eight kilojoules per mole. That is close to the value of about minus sixty six we saw. Right? Now, here is the trap: forgetting the calorimeter. Then q is four hundred and eighteen times three point five, which gives one thousand four hundred and sixty three joules, and the answer becomes minus fifty eight point three kilojoules per mole, about ten per cent too small in magnitude. Got it?

7:33Quick revision: heat equation, sign of delta H, Hess's law

So, quick revision. The heat is q equals m times c plus capital C cal, times delta T, with capital C cal found first from warm and cold water. The enthalpy of solution is minus or plus q over n, negative when the temperature rises. Produce the cooling line back, and use Hess's law for the hydration: anhydrous minus hydrate. Now go and try the practice problems on this page. All the best!