Tag: introduction to equilibrium

Questions Related to introduction to equilibrium

Identify the common property for a chemical reaction at dynamic equilibrium.

  1. The measurable properties like concentration, density, colour, pressure etc remain constant at constant temperature

  2. The forward and backward reactions take place with the same rate

  3. It can be achieved from both directions

  4. All of the above


Correct Option: D
Explanation:

At dynamic equilibrium, the reaction rate of the forward reaction is equal to the reaction rate of the backward reaction. A dynamic equilibrium exists once a reversible reaction ceases to change its ratio of reactants/products, but substances move between the chemicals at an equal rate, meaning there is no net change. It is a particular example of a system in a steady state. Equilibrium is the state of equal, opposite rates, not equal concentrations. The measurable properties like concentration, density, color, pressure etc remain constant at constant temperature.

When equilibrium is attained, the concentration of each of the reactants and products become equal.

  1. True

  2. False

  3. Ambiguous

  4. None of these


Correct Option: A
Explanation:

Equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time.Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and backward reactions are generally not zero, but equal. Thus, there are no net changes in the concentrations of the reactant(s) and product(s).

In reversible reactions concentration of:

  1. reactants decreases with time at equilibrium

  2. products increases with time at equilibrium

  3. reactants decreases and then increases with time at equilibrium

  4. reactants and products are constant at equilibrium


Correct Option: D
Explanation:

In reversible reactions, as equation the rates of forward and backward reactions are same so the concentrations of reactants and products are constant at equilibrium.

Which of the following factors are equal in a reversible chemical reaction at equilibrium?

  1. The number of moles of the reactants and products

  2. The potential energies of the reactants and products

  3. The activation energies of the forward and reverse reactions

  4. The rates of reaction for the forward and reverse reactions

  5. The concentrations of the reactants and products.


Correct Option: D
Explanation:

The rates of reaction for the forward and reverse reactions are equal in a reversible chemical reaction at equilibrium. When the concentrations of reactants and products have become constant, a reaction is said to have reached a point of equilibrium. The consistency of measurable properties such as concentration, color, pressure and density can show a state of equilibrium. The rates of the two opposing reactions have become equal. The amount of products and reactants produced are consistent, and there is no net change.

A reaction continues even after the attainment of equilibrium.
  1. True

  2. False


Correct Option: A
Explanation:

A reaction continues even after the attainment of equilibrium.
The reaction proceeds in forward as well as reverse direction. When equilibrium is attained, it is dynamic in nature. The reactants are converted into products through forward reaction and the products are converted into reactants through reverse reaction.

The equilibrium state can be attained from both sides of the chemical reaction.
  1. True

  2. False


Correct Option: A
Explanation:

The equilibrium state can be attained from both sides of the chemical reaction.
Thus, if initially, the concentration of reactants is much higher than the equilibrium concentration and the concentration of products is much lower than the equilibrium concentration, the reaction will proceed in the forward direction till equilibrium is attained. On the other hand,
if initially, the concentration of reactants is much lower than the equilibrium concentration and the concentration of products is much higher than the equilibrium concentration, the reaction will proceed in the reverse direction till equilibrium is attained.

For hypothetical reversible reaction $\dfrac {1}{2}A _{2}(g) + \dfrac {1}{2}B _{2}(g) \rightarrow AB _{3}(g); \triangle H = -20\ KJ$ if standard entropies of $A _{2}, B _{2}$ and $AB _{3}$ are $60, 40$ and $50\ JK^{-1} mole^{-1}$ respectively. The above reaction will be in equilibrium at

  1. $400\ K$

  2. $500\ K$

  3. $250\ K$

  4. $200\ K$


Correct Option: A

The equilibrium constant for a reaction is $K$, and the reaction quotient is $Q$. For a reaction mixture, the ratio $\dfrac {K}{Q}$ is $0.33$. This means that:

  1. the reaction mixture will equilibrium to form more reactant species

  2. the reaction mixture will equilibrium to form more product species

  3. the equilibrium ratio of reactant to product concentrations will be $3$

  4. the equilibrium ratio of reactant to product concentrations will be $0.33$


Correct Option: A
Explanation:

We know that for a reaction if $\dfrac{K}{Q} < 1,$ the reaction proceeds in backward direction. Hence, the reaction mixture forms more reactant species.

Assume that the decomposition of $H{ NO } _{ 3 }$ can be represented by the following equation
$4H{ NO } _{ 3 }(g)\rightleftharpoons 4{ NO } _{ 2 }(g)+2{ H } _{ 2 }O(g)+{ O } _{ 2 }(g)\quad $'and the reaction approaches equilibrium at $400K$ temperature and $30$ atm pressure. The equilibrium partial pressure of $H{ NO } _{ 3 }$ is $2$ atm
Calculate ${K} _{c}$ in ${ \left( mol/L \right)  }^{ 3 }$
(Use: $R=0.08atm-L/mol-K$)

  1. $4$

  2. $8$

  3. $16$

  4. $32$


Correct Option: A

The optical rotation of the $\alpha-form$ of a pyramose is $+150.7^{\circ}$, that of the $\beta - form$ is $+52.8^{\circ}$. In solution an equilibrium mixture of these anomers has an optical rotation of $+80.2^{\circ}$. The percentage of the $\alpha$ form in equilibrium mixture is:

  1. $28$%

  2. $32$%

  3. $68$%

  4. $72$%


Correct Option: A
Explanation:

$\alpha $ from $=+150.{ 7 }^{ 0 }$, $\beta $ from $=+52.{ 8 }^{ 0 }$

at equilibrium optical rotation $=+80.2$ 
let, at equilibrium $\alpha $-from exist $=x$
      at equilibrium $\beta $-from exist $=(100-x)$
Therefore, $\dfrac { 150.7x+\left( 100-x \right) \times 52.8 }{ 100 } =80.2$
$\Rightarrow \quad 150.7x+5280-52.8x=8020$
$\Rightarrow \quad 99.9x=2740$
$\Rightarrow \quad x=27.42\approx 28$%