Question

Consider the following system at equilibrium where Kc = 1.80×10-2and H° = 10.4 kJ/mol at 698...

Consider the following system at equilibrium where Kc = 1.80×10-2and H° = 10.4 kJ/mol at 698 K.
2 HI (g) -->H2(g) + I2(g)

The production of H2(g) is favored by:

Indicate True (T) or False (F) for each of the following:

___TF 1. increasing the temperature.
___TF 2. increasing the pressure (by changing the volume).
___TF 3. increasing the volume.
___TF 4. removing HI .
___TF 5. adding I2.

Consider the following system at equilibrium where Kc = 55.6 and H° = -10.4 kJ/mol at 698 K.

H2(g) + I2(g) -->2 HI (g)

The production of HI (g) is favored by:

Indicate True (T) or False (F) for each of the following:

___TF 1. increasing the temperature.
___TF 2. decreasing the pressure (by changing the volume).
___TF 3. increasing the volume.
___TF 4. adding HI .
___TF 5. removing I2.

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Answer #1

1)

2HI(g) --------> H2(g) + I2(g)

Abswer

True 1. Increasing the temperatutre

  False 2. Increasing the pressure ( by changing the volume)

False 3. Increasing the volume

False 4. Removing HI

False 5 .Adding I2

Explanation

This reaction is endothermic reaction, endothermic reaction is favored by increase in temperature , so, production of H2 is favored by increasing temperature.

Number moles of gaseous reactants and gasrous products are equal , so change in temperature or change in volume would not affect equilibrium

Removing HI and adding I2 shift the equilibrium to left, so removing HI and adding I2 would not favor production of H2.

2)

H2(g) + I2(g) -------> 2HI(g)

Answer

False 1. Increasing the temperature

False 2 . Decreasing the pessure ( by changing volume)

False 3. Increasing the volume

False 4. Adding HI

False 5. Removing I2

Explanation

Exothermic reaction is defavored by increase in temperature, so increasing temperature would not favor production of HI

Number moles of gaseous reactants and products are equal for these reaction , so decreasing pressure and increasing the volume would not favor production of HI

Adding HI and removing I2 shift the equilibrium to left , so adding HI and removing I2 would not favor production of HI.

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