Could someone please help me on number 4??? and double check my answer for #5. #

Question

Could someone please help me on number 4??? and double check my answer for #5.

#5

a) shifts right

b) shifts left

c) does not change

22 4. Write the competing equilibrium that will occur when a strong acid is added to the following equilibrium: 2 CO , 2 a + 2 H * (aq) = Cr O, Aug ) + H,0am ( 5. Consider this equilibrium system: N2+ 3 H, e = 2 NH , 2 What would happen to the amount of ammonia present for each of the following changes? a Hydrogen is added. b. Nitrogen is removed. C. The volume of the system is increased.

Explanation / Answer

Ans. Le Chatelier’s principle states “if a dynamic equilibrium is disturbed by changing the conditions (Concentration, Volume, Pressure, temperature, etc.), the position of equilibrium shifts to counteract the change to reestablish an equilibrium”.

#4. To present a clear-cut solution, we proceed with hypothetical values (you can choose any value you wish)-

            2 CrO42- + 2 H+ <-------> Cr2O72- + H2O             ; Keq = 100

Equilibrium constant for the reaction, Keq = [Cr2O72-] / ([CrO42-]2 [H+]2)

We reduce the concentrations with A, B and C respectively for demonstration.

            Keq = [A] / ([B]2 [C]2) = 100                     - equation 1

# If you increase C or [H+] by adding strong acid, say new concentration is 5 times that of initial. So, 5C

Now, Keq’ = [A] / ([B]2 [5 C]2) = [A] / ( 25 [B]2 [C]2)      -equation 2

# If Le Chatelier’s principle is followed we have to maintain keq to 100.

What will you do to make Keq = Keq’ ? – to restore equilibrium

What will you do to make [A] / ([B]2 [C]2) = [A] / ( 25 [B]2 [C]2) ? – to restore equilibrium

Solution: If you multiply the numerator of equation 1 by 25, both becomes equal.

So, you need to increase [A] i.e. the product by 25 times. If you do so, the equilibrium will be restored.

Note: Reaction quotient, Q is more suitable for Keq’. Keq’ is used in place of Q for simplicity.

Conclusion: If concertation of a reactant is increased, the reaction must proceed to the right to form more products in order to restore equilibrium.

Answer to #4. The reaction proceeds to the right.

Note that equilibrium constant is a constant value at specified temperature for particular reaction. Change in concertation of any chemical species of the reaction causes left or right shift to the reaction. However, the equilibrium constant remains constant/ unaffected as long as temperature is kept constant.

#5. #a. Reaction proceeds to the right. [Do NOT write “equilibrium shifts to the right” because equilibrium constant remains constant to restore which the reaction must proceed to the right.]

#b. Reaction reverts to the left.

#c. Increasing the volume decrease overall pressure in the vessel. So, the equilibrium constant, Kp also decreases. To counteract decrease in Kp (or, total pressure of the system), more and more gas molecules must be produced.

To produce more gas molecules and restore pressure, the product NH3 must dissociate into N2 and H2. Note that dissociation of 2 mol NH3 forms 4 mol products – so, going to the left increase the number of gas molecules and may restore pressure after some time.

Therefore, the reaction shall revert to the left.

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