Chemistry, 27.04.2020 02:07 larreanathalie3523

A 0.1018 g sample of organic compound with a molar mass of 225.4 g/mol was
burned in the bomb calorimeter with the heat capacity 2.95 kJ/K. The
temperature in the calorimeter increased by 2.63 K. Calculate the energy of
combustion of the compound in kJ/mol.

-1.72 x 10^4
-0.0318
-6.62 x10^1
-1.78 x 10^2
-0.0035

Answers: 3

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Answer from: shyiann7910

Q = -17,200 Kj/mole = -1.72 x 10⁴ Kj/mol

Explanation:

Calorimeter Equation => Q = Cv·ΔT

For the 0.1018 gram sample 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).

Q(for 0.1018g sample) = (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.

Moles of Organic Sample = 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 'increased' temp. by 2.63 K. This indicates the sample, upon burning, was releasing heat energy 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 decreased 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. However, the problem in this post is 'exothermic' and should be accompanied by a negative (-) signage.

Answer from: Quest

a) the particles of the chair are strongly held together.

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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A 0.1018 g sample of organic compound with a molar mass of 225.4 g/mol was
burned in the bomb...

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