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Use the data given below to construct a Born-Haber cycle to determine the electron affinity of Br. ΔH°(kJ) K(s) → K(g) 89 K(g) → K+(g) + e- 419 Br2(l) → 2 Br(g) 193 K(s) + 12Br2(l) → KBr(s) -394 KBr(s) → K+(g) + Br-(g) 674 Use the data given below to construct a Born-Haber cycle to determine the electron affinity of Br. ΔH°(kJ) K(s) → K(g) 89 K(g) → K+(g) + e- 419 Br2(l) → 2 Br(g) 193 K(s) + Br2(l) → KBr(s) -394 KBr(s) → K+(g) + Br-(g) 674 +246 kJ -464 kJ -325 kJ +367 kJ -885 kJ

Respuesta :

ajeigbeibraheem

Answer:

-325 kJ

Explanation:

Data Given:

                                                       ΔH°(kJ)

K(s) → K(g)                                         89

K(g) → K+(g) + e-                               419

Br2(l) → 2 Br(g)                                 193  

K(s) + [tex]\frac{1}{2}[/tex] Br[tex]_2[/tex](l) → KBr(s)                       -394

KBr(s) → K+(g) + Br-(g)                     674

Let take this process one after the other ;

To start with; in the first reaction:

We see the conversion of potassium(s) changing to potassium (g); this change in process is called Sublimation. The ΔH[tex]_{sub}[/tex] = 89 kJ

In the second reaction; There is a removal of electron taking place in the ionization potential.  So, the ΔH[tex]_{Ip}[/tex] = 419 kJ

For the third reaction; dissociation of bromine occurs ; whereby the dissociating energy is the heat of reaction.

We were given for two bromine but we require only for just one bromine; so  [tex]\frac{1}{2} *193[/tex] =  96.5 kJ

In the fourth reaction;

Br (g) + e- ----> Br- (g)

here the electron affinity (ΔH[tex]_{EA[/tex]) = ???

In the fifth reaction: The reaction is given in  reserve indicating it is exothermic and also the reaction signifies the heat of crystallization. So  ΔH[tex]_{cry}[/tex] = - 674 kJ

The final reaction   K(s) + [tex]\frac{1}{2}[/tex] Br[tex]_2[/tex](l) → KBr(s) shows the heat of formation with the ΔH[tex]_{for[/tex] = -394 kJ

We all know that ; the heat of formation is equal to the sum of all other enthalpies, therefore:

[tex]\Delta H_{for} = \Delta H_{sub} + \Delta H_{ip} + \Delta H_{DE}+\Delta H_{EA} + \Delta H_{cry}[/tex]

-394 = 89 + 419 + 96.5 + ΔH[tex]_{EA[/tex] +( - 674)

- ΔH[tex]_{EA[/tex] = -96.5 + 394

- ΔH[tex]_{EA[/tex] = 324.5

ΔH[tex]_{EA[/tex] = -324.5

Therefore ; the electron affinity for Br is  ≅ -325 kJ

The branch of science which deals with chemicals is called chemistry.

The correct answer is 325 kJ

The data is given in the question is as follows:-

Reaction                                            ΔH°(kJ)

  • K(s) --- >K(g)                                          89
  • K(g) --->K+(g) + e-                                 419
  • Br2(l) ----> 2 Br(g)                                  193  
  • K(s) +  Br(l) ---> KBr(s)                           -394
  • KBr(s) ----> K+(g) + Br-(g)                        674

The conversion of solid into gas is called sublimation. The  energy required in the process is ΔH = 89 kJ

In second reaction, the pottassium removes the electrons So, the energy required in the process is ΔH = 419 kJ

  • The energy of the bromine is  96.5 kJ

we have to find the electron affinity.

The exothermic energy in the fourth reaction is -674 kJ

The energy in the fifth reaction shows the heat of formation with the is 394 kJ.

After calculation all the enthalpy we will get:-

[tex]-394 = 89 + 419 + 96.5 + H +( - 674)[/tex]

After solving the equation the heat energy is as follows:-

  • [tex]H = -324.5[/tex]

Hence, the correct option is -325 kJ

For more information, refer to the link:-

https://brainly.com/question/25026730

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