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3 years ago

PLEASE HELP I HAVE THIS TEST TOMORROW AM GIVING MORE POINTS THAN I SHOULD!

Chemistry

1 answer:

BlackZzzverrR [31]3 years ago

8 0

Answer:

match emit 350 J light energy.

Explanation:

By law of conservation of energy, total energy before and after reaction for an isolated system remains constant.

Total energy for unlit match = 1000 j

When it burns type of energy released

thermal energy = 400 j

light energy --> it value is unknown which we have to find

also remaining match has 250 joule of chemical energy

applying law of conservation of energy

energy before match was unlit = energy after match was lit

1000 j = thermal energy + light energy + chemical energy of remaining match

1000 J = 400 J + light energy + 250 J

light energy = 1000 J - 400J - 250J

light energy = 350 J

Thus, match emit 350 J light energy.

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Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

I hope it helps you!

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