Chapter 7: Equilibrium

Q51Written answer

How can you predict the following stages of a reaction by comparing the value of ${{\mathbf{K}}_{\mathbf{c}}}\text{ }\mathbf{and}\text{ }{{\mathbf{Q}}_{\mathbf{c}}}$?

(i) Net reaction proceeds in the forward direction.

(ii) Net reaction proceeds in the backward direction.

(iii) No net reaction occurs.

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Ans: The values of ${{\text{K}}_{\text{c}}}$ and ${{\text{Q}}_{\text{c}}}$ are itself sufficient to explain the direction of reaction and less than or greater than one another decides the direction in which reaction will proceed as follows-

(i) As ${{\text{Q}}_{\text{c}}}$ < ${{\text{K}}_{\text{c}}}$, the reaction proceeds in the forward direction.

(ii) If ${{\text{Q}}_{\text{c}}}$ > ${{\text{K}}_{\text{c}}}$, the reaction will proceed in the direction of reactants (reverse reaction).

(iii) If ${{\text{Q}}_{\text{c}}}$ = ${{\text{K}}_{\text{c}}}$ , no net reaction occurs.

Q52Written answer

On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction. ${{\mathbf{N}}_{\mathbf{2}}}\left( \mathbf{g} \right)\text{ }+\text{ }\mathbf{3}{{\mathbf{H}}_{\mathbf{2}}}\left( \mathbf{g} \right)\rightleftharpoons \mathbf{2N}{{\mathbf{H}}_{\mathbf{3}}}\left( \mathbf{g} \right)\text{ }\Delta \mathbf{H}\text{ }=\text{ }\text{ }\mathbf{92}.\mathbf{38}\text{ }\mathbf{kJ}\text{ }\mathbf{mo}{{\mathbf{l}}^{\mathbf{1}}}$. What will be the effect of addition of argon to the above reaction mixture at constant volume?

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Ans: According to Le Chatelier’s principle, when we raise the temperature, it shifts the equilibrium to left and decreases the equilibrium concentration of ammonia since it is an exothermic reaction. In other words, low temperature and high pressure is favourable for high yield of ammonia. There will be no change in equilibria on addition of argon (Ar).

Q53Written answer

A sparingly soluble salt having general formula $\mathbf{A}{{\mathbf{ }}_{\mathbf{x}}}^{\mathbf{p+}}\mathbf{ }{{\mathbf{B}}_{\mathbf{y }}}^{\mathbf{q - }}$. and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.

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Ans: The equation can be written as-

\[ A_{x}^{p+} B_{y}^{q-} \rightleftarrows x A^{p+} (aq) + y B^{q-} (aq) \]

S moles of A and B dissolves to give x S moles of Ap+ and y Bq-.

\[ K_{sp} = [A^{p+}][Bq^-]^{y} = [x^{5}]^x [y^5]^y \]

\[= x^x y^y 5^{x+y} \]

Q54Written answer

Write a relation between ∆G and Q and define the meaning of each term and answer the following:

(a) Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K.

(b) Explain the effect of increase in pressure in terms of reaction quotient Q. for the reaction: $\mathbf{CO}\text{ }\left( \mathbf{g} \right)\text{ }+\mathbf{3}{{\mathbf{H}}_{\mathbf{2}}}\text{ }\left( \mathbf{g} \right)\to \mathbf{C}{{\mathbf{H}}_{\mathbf{4}}}\text{ }\left( \mathbf{g} \right)+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\text{ }(\mathbf{g})$.

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Ans: (a) As we know,

ΔG= $\text{ }\!\!\Delta\!\!\text{ }{{\text{G}}^{\odot }}$+RTlnQ

$\text{ }\!\!\Delta\!\!\text{ }{{\text{G}}^{\odot }}$= Change in free energy as the reaction proceeds,

ΔG = Standard free energy change,

Q = Reaction quotient,

R = Gas constant,

T = Absolute temperature.

Since, $\text{ }\!\!\Delta\!\!\text{ }{{\text{G}}^{\odot }}$ =−RTlnK

∴ΔG=−RTlnK+RTlnQ=RTlnKQ

If Q<K, ΔG will be negative. Reaction proceeds in the forward.

If Q=K, ΔG=0, reaction is in equilibrium and no net reaction is there.

Ans: When we increase the pressure equilibrium will shift in forward direction it means Q<K.

Types of Chemical Reactions :

There are 4 main types of chemical reactions. These are-

1. Combination Reactions- Combination Reactions are chemical reactions with 2 or more substances combining to form one single compound.

2. Decomposition Reactions- Decomposition Reactions are chemical reactions with one compound decomposing to produce 2 or more different substances. The digestion process of food in our body is a Decomposition reaction.

3. Displacement Reactions- Displacement reactions are reactions that involve the displacement of 1 element or group by another.

4. Double Displacement- Displacement reactions are reactions with 2 compounds that react to form 2 new compounds. In the displacement reaction, no change in oxidation state occurs.

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