Equilibrium

Answer (d)

Those compound which accept is called Bronsted base
accept and form . So it is a Bronsted base.

or
Now, (given)

At T > 15 K
RT > 0.0821 15 L atm mol
RT > 1.22 L atm mol

No. of moles of gaseous products > No. of moles of gaseous reactants
Such a reaction in the forward direction will be favoured by low pressure.

The reaction taking place in a lime kiln is

Removing one of the product will shift the equilibrium in the forward direction. As such removing as and when it forms, will shift the equilibrium in the forward direction. This will increase the yield of

Answer (c) Increasing the temperature

(Henderson's equation)

Number of gaseous product

- Number of gaseous reactants

Total number of moles at equilibrium

Number of moles of

in the diluted base

Total of water

Total milliequivalents of


Total milliequivalents of


Milli equivalence of left

In the reversible reaction,




Since, the randomness increases (because solid is changing into gas), entropy will increase and thus, is positive. Reversible reaction never undergo to completion (ie, never be instantaneous).

For the given reaction, and we know that




ie, (equilibrium constant) is greater than one.

Moreover, the reaction is thermodynamically feasible.

Relationship between and equilibrium constant is



When is negative.

...(i)

Equation (i) is reversed and multiplied by 2

...(i)
(ii)

Milliequivalents of


Milliequivalents of

Milliequivalents of (base)

milliequivaletns of (acid)

Resulting solution will be neutral.

Its

Milliequivalents of of (a) of distilled water milliequivalents for (b) Milliequivalents of (c) of Milliequivalents (d) Milliequivalents of Milliequivalents of of having is more than that of of , therefore adding this, of solution, does not alter.

Acetic acid and sodium acetate, citric acid and sodium citrate are acidic buffers, whereas ammonium hydroxide and ammonium chloride is a basic buffer solution. A neutral buffer is a solution of weak acid (acetic acid) and weak base (ammonium hydroxide).

Sulphides of group-II are precipitated by passing gas through the solutions of cations in the presence of dilute HCl.

being a weak electrolyte ionises only slightly while HCl being a strong electrolyte is almost completely ionised.

.

Thus, the common ion effect takes place. As a result, the degree of dissociation of H2S decreases sufficiently and the concentration of ions in the solution becomes very small, which results in the precipitation of those sulphides having very low solubility product.

Given relations




On comparing Eqs. (i) and (ii) and to make

E (i) = E (ii),

Reverse E (i) and multiply it with 2 , we get



Thus, E (ii) = E (iii) and,

Qn adding catalyst. the equilibrium attain 10 time factor.

As equilibrium constant depends on temperature. It will not change, however the reaction attains equilibrium 10 times faster hence

The direction in which the reaction is expected to proceed

(reaction goes from right to left) In this case, the reaction proceeds in the reverse or the backward direction or in the direction of the reactants.

(the reaction goes from left to right) In this case, the reaction proceeds in the forward direction or the direction of the products.

(reaction is at equilibrium) In this case, the reaction is said to be at equilibrium.

But the reaction is dynamic in nature.

in is electron deficient and hence, it has a tendency to accept a pair of electrons.

= number of moles of gaseous products -number of moles of gaseous reactants

Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will shift to counteract that change. In this case:
- Increasing the concentration of CO will shift the reaction to the right, favouring the formation of CH_4.
- Decreasing the concentration of CH_4 will also shift the equilibrium to the right to produce more CH_4.
Thus, increasing the concentration of CO and decreasing the concentration of CH_4 favours the formation of methane.

The equilibrium expression for the reaction is: At equilibrium, if the initial concentrations of all species are 1 M, we assume the change in concentration is . Therefore, the concentrations at equilibrium are: - [A] = 1 - x - [B] = 1 - x - [C] = 1 + x - [D] = 1 + x Substitute these values into the equilibrium expression: Solving for , we get: Simplifying: Expanding both sides and solving for , we find . Thus, the equilibrium concentration of D is:

Statement (a) is true because equilibrium can only be established in a closed system, where the system’s components do not escape.
Statement (b) is true because at equilibrium, properties like concentration, pressure, and temperature remain constant over time.
Statement (c) is true as well. The equilibrium constant for the reverse reaction is indeed the inverse of the equilibrium constant for the forward reaction, i.e., .

Step 1: Understanding Le Chatelier's Principle.
Le Chatelier's principle states that if a change of condition (like temperature, pressure, or concentration) is applied to a system in equilibrium, the system will shift in a direction that counteracts the change. We need to see how temperature and pressure changes will affect the equilibrium position.
Step 2: Effect of Temperature.
The reaction is given as exothermic ( H = -q kJ, meaning heat is released). According to Le Chatelier's principle:

• Increasing the temperature will favor the endothermic (reverse) reaction to absorb the added heat. The equilibrium will shift to the left, decreasing the amount of product A B.
• Decreasing the temperature will favor the exothermic (forward) reaction to release heat. The equilibrium will shift to the right, increasing the amount of product A B.

Since temperature changes the equilibrium position, the amount of product is affected by temperature.
Step 3: Effect of Pressure.
The effect of pressure depends on the change in the number of moles of gas ( ). Since , pressure will affect the equilibrium. According to Le Chatelier's principle:

• Increasing the pressure will favor the side with fewer moles of gas to relieve the pressure. The equilibrium will shift to the right, increasing the amount of product A B.
• Decreasing the pressure will favor the side with more moles of gas. The equilibrium will shift to the left, decreasing the amount of product A B.

Since pressure changes the equilibrium position, the amount of product is affected by pressure.
Step 4: Effect of a Catalyst.
A catalyst increases the rate of both the forward and reverse reactions equally. It helps the system reach equilibrium faster but does not change the position of the equilibrium. Therefore, a catalyst does not affect the amount of product at equilibrium.
Step 5: Final Answer.
The amount of product is affected by both temperature and pressure, but not by a catalyst. Therefore, option (C) is the best answer.