1) Consider the following system at equilibrium where K c = 55.6 and deltaH° = -

Question

1) Consider the following system at equilibrium where Kc = 55.6 and deltaH° = -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:

___T/F 1. decreasing the temperature.
___T/F 2. increasing the pressure (by changing the volume).
___T/F 3. increasing the volume.
___T/F 4. removing HI .
___T/F 5. removing I2

2) Consider the following system at equilibrium where Kc = 1.80×10-4and delta H° = 92.7 kJ/mol at 298 K.
NH4HS (s) ---->NH3(g) + H2S (g)
The production of NH3(g) is favored by:

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

___T/F 1. increasing the temperature.
___T/F 2. increasing the pressure (by changing the volume).
___T/F 3. increasing the volume.
___T/F 4. adding NH4HS .
___T/F 5. removing H2S .

Explanation / Answer

Ans 1

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

H° = -10.4 kJ/mol - - - - exothermic reaction

From the Le Chatelier's principle

Statement 1 is true

The reaction is exothermic, decrease in the temperature will favor the forward reaction or the production of HI

If the reaction is endothermic, increase in the temperature will favor the forward reaction or the production of HI

Statement 2 is false.

Moles of reactants = moles of products

So, there is no change in pressure and no change in volume

Statement 3 is also false as same as statement 2

Statement 4 is true

Decrease in concentration of products or removing HI will lead to forward reaction, equilibrium shift towards the formation of HI, more HI will be formed.

Statement 5 is false

Removing I2 means

Concentration of reactants < concentration of products

The reaction will be in the reverse direction and less HI will be formed.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.