Solution
The reaction is an endothermic
reaction. Heat is absorbed as the temperature decreases.
The calibration factor of the calorimeter is given by the expression below.
Obtain the amount of energy absorbed from 0.996 grams of ammonium nitrate
Calibration factor X change in temperature = 468 X 0.672 = 0.314KJ
The enthalpy change is energy/mol
SO!
Step 1) Mole of ammonium nitrate added = 0.996 /80 = 0.0125 mole
Step 2) Divide the amount of energy by the number of mole of ammonium nitrate.
0.314/ 0.01245 = 25.2Kj/mol
Note how the enthalpy change has a positive sign. This indicates an endothermic reaction (heat absorbed from the surroundings).
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Calorimetry
exercise 3 |
A current of 2.73 A was passed through the heating coil of a calorimeter for one minute at a potential difference of 6.00 V. The temperature increased from 20.340 to 22.440 degrees Celsius. N =14, C = 12, O = 16, H =1, K =39 |
0.996 grams of ammonium nitrate was dissolved in a calorimeter and the temperature dropped from 20.34 to 19.668. a) Is this an exothermic or endothermic reaction? Explain. b) Write the equation for the reaction. c) Calculate the enthalpy change of the equation in b).
Solution |