Answer:
Q = -17,200 Kj/mole = -1.72 x 10⁴ Kj/mol
Explanation:
Calorimeter Equation => Q = Cv·ΔT
For the <u>0.1018 gram sample</u> of organic compound burned in calorimeter ...
Cv = calorimeter heat capacity at constant volume = 2.95 Kj/K (Cv is typically supplied by manufacturer of the calorimeter as a standard for the instrument).
ΔT = temperature change experienced by calorimeter water in Kelvin (K) values = +2.63 K (NOTE => increasing (+) temp indicates heat flow from sample to calorimeter water => Exothermic Rxn).
<u>Q(for 0.1018g sample)</u> = (2.95 Kj/K)(2.63 K) = 7.7585 Kj
Converting to Kj/mole => divide the sample heat value (7.7585 Kj) by moles of sample used => moles of sample = mass/molar mass.
<u>Moles of Organic Sample</u> = 0.1018g/225.4g·mol⁻¹ = 4.52x10⁻⁴ mole
∴ Q(kj/mole) = Q/n = 7.7585 Kj/4.52x10⁻⁴mole = 1.72x10⁴ Kj/mole (exothermic)*
*Why 'exothermic'? => Since the reaction was carried out in a bomb calorimeter the energy flow of the reaction is measured in terms of a temperature change. The problem dialogue indicates <em>'increased' </em>temp. by 2.63 K. This indicates the sample, upon burning, was <u>releasing heat energy</u> into the surroundings. That is, the heat flow is toward the water jacket of calorimeter and away from the sample thereby heating the water and showing an increase in temperature. This type of heat flow is 'exothermic' with respect to the system sample and it's energy value is posted with a 'negative' sign to indicate heat loss during process. Therefore, the molar heat of combustion = -17,200 Kj/mole.
NOTE: Endothermic Reactions
For a notation in problems that the temperature measurement is discribed as <u><em>decreased</em></u> would indicate an endothermic reaction - heat flow from surroundings toward sample of interest. In such a case, a plus (+) signage should accompany the Q-value calculated. <em>However, the problem in this post is 'exothermic' and should be accompanied by a negative (-) signage.</em>
<em />
