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MariettaO [177]
3 years ago
9

What is the significance of "Er" in the diagram?

Chemistry
2 answers:
vladimir2022 [97]3 years ago
8 0

Answer:

The significance of "Er" in the diagram is :

B.) Threshold energy for reaction

Explanation:

Threshold energy : It is total amount of energy required by the reactant molecule to reach the transition state .

Activation energy : It is the excess energy absorbed by the molecules to reach the transition state.

Activation Energy = Threshold Energy - Average Kinetic Energy

This means Activation energy decreases on increasing kinetic energy

On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.

Catalyst also reduces the Activation energy.

Explanation:

Drupady [299]3 years ago
3 0

Answer:

The significance of "Er" in the diagram is :

B.) Threshold energy for reaction

Explanation:

Threshold energy : It is total amount of energy required by the reactant molecule to reach the transition state .

Activation energy : It is the excess energy absorbed by the molecules to reach the transition state.

<u>Activation Energy = Threshold Energy - Average Kinetic Energy</u>

<u>This means Activation energy decreases on increasing kinetic energy</u>

On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.

Catalyst also reduces the Activation energy.

<u>Er = Threshshold energy for reaction at 30 degree</u>

<u>Ea = Activation Energy</u>

<u>The given figure shows that the threshold energy decreases on increasing the temperature</u>

<u>Only the molecule having energy greater than Er can react to form product</u>

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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.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

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