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podryga [215]
4 years ago
6

When 1.550 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87∘C to 38.13∘C.

Find ΔErxn for the reaction in kJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘C. Express your answer in kilojoules per mole to three significant figures.2. The combustion of toluene has a ΔErxn of –3.91×103 kJ/mol. When 1.55 g of toluene (C7H8) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12∘C to 37.57∘C. Find the heat capacity of the bomb calorimeter. Express the heat capacity in kilojoules per degree Celsius to three significant
Chemistry
1 answer:
AlladinOne [14]4 years ago
3 0

Explanation:

1). The given data is as follows.

       T_{i} = 25.87^{o}C,      T_{f} = 38.13^{o}C

               C = 5.73 kJ/^{o}C

Hence, calculate the change in enthalpy of the reaction as follows.

      \Delta E_{rxn} = -C \times \Delta T

                     = -5.73 \times (38.13 - 25.87)^{o}C

                     = -70.25 KJ

As,  number of moles = \frac{mass}{\text{molar mass}}

                                    = \frac{1.55}{(6 \times 12 + 14 \times 1)}

                                    = 0.018 mol

Therefore, enthalpy of reaction in kJ/mol hexane is as follows.

            \Delta E_{rxn} = \frac{-70.25 KJ}{0.018 mol}

                          = -3.90 \times 10^{3} kJ/mol

Thus, we can conclude that \Delta E_{rxn} for the reaction in kJ/mol hexane is -3.90 \times 10^{3} kJ/mol .

2).  As we know that,

     Number of moles = \frac{mass}{\text{molar mass}}

                                  = \frac{1.55}{(7 \times 12 + 8 \times 1)}

                                   = 0.017 mol

       \Delta E_{rxn} = \E_{rxn} per mol \times \text{number of moles}

                      = -3.91 \times 10^{3} \times 0.017

                      = -65.875 kJ

As,    \Delta E_{rxn} = C \times \Delta T


                    -65.875 kJ = -C \times (37.57 - 23.12)^{o}C

                           C = 4.56 kJ/^{o}C

Hence, heat capacity of the bomb calorimeter is 4.56 kJ/^{o}C.

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Answer:

Here's what I get  

Explanation:

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