Calorimetric Calculations - A lvl Chem

Lord dejavu SG

Hi all. This is regarding calorimetric calculations involoving the use of Q = m x c x /_\ T (delta T) / n.

1) 100 cm^{3 of 0.5 mol/dm}3 silver nitrate solution and 100cm^{3 of 0.5 mol/dm}3 potassium chloride solution were placed in seperate plastic bottles and stirred. Both solutions were intially at the same temperature. After mixing, the temperature of the mixture rise by 7.5 degree celcius. Calculate the standard enthalpy change of reaction. Ans: (-1.25×10^{5) J/mol}

For this question, i gathered that delta T is (7.5 + 273)K. c is as usual 4.18 and the mass of the solution is (100 + 20) g. Am i right so far?

My problem is finding the amount. How do i know which amount to use? Add both the amount of silver nitrate and potassium chloride or use limiting reagant?

2)When 60cm3 of 0.1 mol/dm^{3 sulphuric acid and 40cm}3 of 0.20 mol/dm^{3 sodium hydroxide were mixed in a calorimeter,
the temperature rose by 6.5 degree celcius. Calculate the standard
enthalpy change of neutralisation.

Ans: (-3.41×10}5 J/mol)

delta T in this case is (6.5 + 273) K, c is 4.2 J/g/K, and the mass is (60cm^{3 + 40 cm}3)

Same problem as qn 1, i dont know how to find the amount ( of water since neutralisation). I used limiting reagant as the amount of sulphuric acid (0.006) and sodium hydroxide (0.00 Cool does not tally with the theroetical mole ratio of 2:1.

Hence, the amount of water i use is 0.016 mol but my answer is wrong.

Thanks guys (:

UltimaOnline SG

Magnitude of change of temperature in Kelvins is the same as that of change in Celsius. In other words, delta T = delta t (where T = Kelvins, t = Celsius).

Do not add 273. You want the magnitude of change – that’s the meaning of delta; “delta T” means “change in temperature” – and not the absolute temperature in Kelvins.

UltimaOnline SG

For both questions, use no. of moles of limiting reactant when calculating enthalpy of reaction. Ignore the excess reactant (but include its volume of solution heated up by the exothermic enthaply change) in this regard.

You should use ionic equations rather than chemical equations.

Ag+(aq) + Cl-(aq) -> AgCl(s)

and

H+(aq) + OH-(aq) -> H2O

Lord dejavu SG

Thanks UltimaOnline! I managed to solve both questions using the ionic equtions. But how do i know when to use the ionic or chemical equation? In some instances, the chemical equation allows me to find which reactant is the limiting reagant. Smile

Lord dejavu SG

Another question:

(HCO3)2- (aq) + H+ (aq) —> H2O (l) + CO2 (g)
To determine the standard enthalpy, equal volumes of 40.0cm each of 1.0 mol dm^{3 (HCO3)2 (aq) and 0.5 mol dm}-3 H+ (aq) were mixed.

A change in temperature of -0.5C was recorded. Assuming that the process only 90% efficient, calculate the standard enthalpy change of the reaction.

I’ve actually got the solution from my tutor in school.
After apply mc(deltaT) which gives me a value of 185.5 J, she multiplied it by 100/90 before proceeding to solve the question as per normal. Why did she do this? (I always thought that information on the process being 90% efficient is only used after finding the standard enthalpy change. ie take the value of deltaH x 90/100).

UltimaOnline SG

For most questions, it doesn’t matter whether you use chemical or ionic equation, as long as you write your equations (whether ionic or chemical) correctly, you will end up with the same correct answer.

Consider the 1st question below. If you choose to use chemical equation, you will have to take more care in balancing the chemical equation, but you can apply stoichiometry directly. If you choose to write ionic equation instead, the immediate stoichiometry is simplier (1 proton combines with 1 hydroxide ion), but you will have to do a further calculation to relate protons or hydroxide ions to the acid or base. Eg. If molarity of protons = 0.9 mol/dm3, then molarity of triprotic phosphoric(V) acid must be 0.9/3 = 0.3 mol/dm3.

There are some cases, in which you have to use ionic equation instead. Eg. if there are 2 or more acids involved. From the amount of base you use, calculate the amount of hydroxide ions involved. Hence apply 1:1 stoichiometry and state the amount of protons involved. Then apply further calculations to determine how much of the protons came from one acid, and how much came from the other.

Similarly, in the 2nd question below, you have to determine the total no. of moles of ethandioate ions present (from both the acid and the salt), by applying stoichiometry against KMnO4. Meaning, in such questions, using the ionic (rather than chemical) equation makes more sense.

The 2 questions which follow are considered challenging for ‘O’ levels (even ‘A’ level students struggle with them), so don’t attempt them unless you’re sure of yourself. I will not be posting detailed worked solutions to these questions, only the final answers are given below.

Q1. Acid-Base stoichiometry.
125cm3 of a solution of phosphoric(V) acid were mixed with 75cm3 of 0.25mol/dm3 of barium hydroxide. A 50cm3 aliquot of the resultant mixture required 7.5g of sodium carbonate for exact neutralization, in which a gas was produced. By calculating the molarity of phosphoric(V) acid used, calculate the sample mass of solid phosphoric(V) acid that needs to be dissolved in 125cm3 of pure water to prepare this molarity.
Ans :
The sample mass of solid phosphoric(V) acid that needs to be dissolved is 19.7g

Q2. Redox + Acid-Base stoichiometry.
A solution contains a mixture of anhydrous ethanedioc acid, and sodium ethanedioate. 25 cm3 of this solution required 20.0 cm3 of 0.10 mol/dm3 sodium hydroxide for neutralization using phenolphthalein as indicator, and 28.0 cm3 of 0.02 mol/dm3 of potassium manganate(VII) solution for complete oxidation at 60°C in the presence of excess sulfuric acid. Calculate the molarity of the acid and the salt.
Ans :
0.04 mol/dm3
0.016 mol/dm3

UltimaOnline SG

First, note that HCO3- has a uninegative anionic charge, not dinegative.

In regards to the 90% efficiency :

Heat transferred
= m x c x dT
= 80g x 4.18×0.5
= 167.2 J

Since the process is only 90% efficient, actual heat released = (100/90) x 167.2 = 185.8 J

Enthalpy of reaction
= heat transferred / no of mols
= 185.8 / 0.02
= + 9290 J

(Endothermic, since temperature went down. For ‘A’ level students, try drawing the mechanism of the reaction leading to the liberation of carbon dioxide.)

Lord dejavu SG

Thanks again UltimaOnline, point noted about chemical vs ionic equations.

I still cant get why for actual heat release in the last question, we have to use 100/90×167.2. Why not 90/100×167.2 instead?

UltimaOnline SG

- I still cant get why for actual heat release in the last question, we have to use 100/90×167.2. Why not 90/100×167.2 instead? -

Because the heat change that you measured, is a bit less (ie. 90/100) than the true heat change on which you must base your calculations on.

Because of efficiency problems (ie. heat loss to surroundings), the true heat change is actually a bit more (ie. 100/90) than what your thermometer recorded. So to calculate (true) enthapy change, you need to base your calculations on the (true) heat change, (ie. 100/90).