# Enthalpy of formation of calcium carbonate

F.6 Chemistry/TAS 6/P.()

EXPERIMENT 6                Enthalpy of formation of calcium carbonate

## Objective

To determine the enthalpy of formation of calcium carbonate

Procedures

1. ##### Reaction of calcium with dilute hydrochloric acid

1.        1.0909 g of calcium metal was weighed out accurately.

2.        100 cm3 of approximately 1 M hydrochloric acid was pipetted.and placed in a plastic beaker.

1. The initial temperature of the acid was determined
2. The weighed calcium was added into the acid and stirred thoroughly with the thermometer until all the metal had reacted.
3. The maximum temperature attained by the solution was recorded.
4. The experiment was repeated with 1.0538g calcium metal.

Results:

Calculations and Discussion:

1. What does the term “heat of formation” of a substance mean?
Heat of formation refers to the heat change when one mole of a substance is formed from its constituent elements is their standard states under standard conditions.
2. What are “standard conditions” for thermochemical calculations?
Standard conditions is defined as elements or compounds appear in their normal physical states at a pressure of 1 atm (101325 Nm-2/760mmHg) and at temperature of 25 oC (298 K).Moreover, the solution should have unit activity(1mol dm-3 ).
3. Write the equation for the formation of calcium carbonate under standard conditions.

(Call this Equation 1)

Ca(s) + C(s) + 3/2 O2(g)  → CaCO3(s)

1. Write an ionic equation for the reaction taken place. (Call this Equation 2)
Ca(s) + 2H+(aq.)  Ca2+(aq.) + H2(g)

1. Assuming        (a)        the solution in the plastic beaker has the same specific heat capacity as    water, i.e., 4.2 kJg-1K-1 and (b)        density of the acid is the same as that of water, i.e., 1.0 g cm-3.  Calculate, in each experiment, the heat change in the reaction between the calcium and the acid per gram of calcium.

For the first experiment:

∵Energy evolved by the reaction= Energy absorbed by the acid

∴By E = mc△T,

△H per gram of calcium
= [(100/1000)(4200)(28)]/ 1.0909
= -10780J g
-1
= -10.780kJ g-1

For the second experiment:

∵Energy evolved by the reaction= Energy absorbed by the acid

∴By E = mc△T,

△H per gram of calcium = [(100/1000)(4200)(26)]/ 1.0538
= -10362J g
-1
= -10.362kJ g-1

1. Calculate the average heat evolved by one mole of calcium.
For the first experiment:
△H per one mole of calcium
=[(100/1000)(4200)(28)]/ [1.0909/40.08]
=-432066 J mol
-1
=-432.066 kJ mol-1

For the second experiment:
△H per one mole of calcium
=[(100/1000)(4200)(26)]/ [1.0538/40.08]
=-415329 J mol
-1
=-415.329 kJ mol-1

Average △H per one mole of calcium  (ΔHx)
=(432.066 +415.329)/2
=-423.698 kJ ...

#### Here's what a star student thought of this essay

All punctuation and grammar seems correct but sometimes one or two words have incorrect endings or the 's' missed off. Spelling apart from this seems fine.

The mass of calcium and how it is weighed out is not described very well, and the candidate may have included the mass of the weighing boat in the mass of calcium used. The analysis used is to a high level and all the calculations and scientific definitions are correct. The candidate could have improved their grade by calculating percentage errors in the experiment for the equipment used. References should have been correctly cited throughout the text. Scientific terms used are to a good standard.

Good scientific essay. The response to the different questions are adequate and presented to a high level of scientific correctness. Definitely could be used as a clear and concise building block for anyone looking for ideas about calcium carbonate combustion.