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Equilibrium Test - 18

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Equilibrium Test - 18
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  • Question 1
    1 / -0
    Fear or excitement generally causes one to breathe rapidly and it results in the decrease of concentration of $${CO}_{2}$$ in blood. In what way, it will change the $$pH$$ of blood?
    Solution
    As we know, $$CO_{2}$$ is acidic in nature. So, a decrease in its concentration should increase the $$pH$$. But, blood contains serum which has proteins in it. This acts as a buffer solution. We already know that buffer solutions resist a change in the $$pH$$ even on addition of a small amount of acid or small amount of base or dilution, etc.

    Therefore, the $$pH$$ of the blood remains unaffected.

    Hence,option C is correct.
  • Question 2
    1 / -0
    Indicate the correct answer out of the following for the reaction,
                        $$N{H}_{4}Cl + {H}_{2}O \rightleftharpoons N{H}_{4}OH + HCl$$
    Solution
    Option (C) is  correct.
    The reaction is retarded by the addition of hydrogen ion as increase $$H^{+}$$ favors the backward reaction.
  • Question 3
    1 / -0
    When $$NaOH$$ is dissolved in water, heat is evolved. If the temperature is raised, the solubility of $$NaOH$$:
    Solution
    Solubility is defined as the concentration of the solute in a saturated solution. So when considering the increase in solubility with temperature, you have to check the enthalpy of solution of NaOH in a saturated solution of NaOH.

    When you dissolve NaOH in pure water, the process is highly exothermic, but as the concentration of NaOH increases, the process becomes less exothermic and eventually near the saturation point it becomes endothermic (it becomes less favourable for more NaOH to dissolve).

    So according to Le Chatelier’s Principle, although the dissolution process is overall exothermic, since solubility is determined only at the saturation point, the solubility increases with increase in temperature.
  • Question 4
    1 / -0
    Predict the conditions for forward reaction on the basis of Le Chatelier's principle for :
                $$2S{O}_{2}\left(g\right) + {O}_{2}\left(g\right) \rightleftharpoons 2S{O}_{3}\left(g\right)$$;    $$\Delta H = -198  kJ$$
    Solution
    Option (A) is correct.
    An increase in temperature favours endothermic reaction whereas an increase in pressure favours the reaction showing a decrease in mole or volume.
    An increase in temperature is the result of the flow of heat into the system. In an endothermic reaction, an increase in temperature favours the reaction to occur in the forward direction reverse in case of exothermic. 

    When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.
  • Question 5
    1 / -0
    If pressure is applied to the equilibrium of solid $$\rightleftharpoons$$ liquid. The melting point of the solid :
    Solution

    If pressure is applied to the equilibrium of solid $$\rightleftharpoons$$ liquid. The melting point of the solid may increase or decrease depending upon the nature of solid.

    As, $$V\left(s\right) > V\left(l\right)$$ for ice-water and $$V\left(s\right) <V\left(l\right)$$ for others.

    For most substances, increasing the pressure when a system is in equilibrium between liquid and solid phases will increase the phase transition temperature. Water is one of a few special substances for which the pressure lowers the temperature of transition. The basic reason is that water actually expands when it goes from the liquid to solid phase.

  • Question 6
    1 / -0
    According to Le Chatelier's principle adding heat to a solid and liquid in equilibrium with endothermic nature will cause the :
    Solution
    According to Le Chatelier's principle adding heat to a solid and liquid in equilibrium with endothermic nature will cause the mass of solid to decrease.

    Solid $$\rightleftharpoons$$ Liquid;  $$\Delta H = +ve$$;

    Increase in temperature favours forward reaction. In endothermic reactions, ( $$\Delta H = +ve$$) heat is absorbed with the reactants. It means more energy is needed to overcome the forces of attraction between molecules and to separate them from one another than the energy gained when new bonds are formed.
  • Question 7
    1 / -0
    A characteristic feature of reversible reactions is that:
    Solution
    A characteristic feature of reversible reactions is that they never proceed to completion.
    Reversible reactions always attain equilibrium and never go for completion. Reversible reactions, however a static equilibrium is not being established.In these most of the reactions when carried out in closed vessels do not go to completion, under a given set of conditions of temperature and pressure. In fact in all such cases, in the initial state, only the reactants are present but as the reaction proceeds, the concentration of reactants decreases and that of products increases. Finally a stage is reached when no further change in concentration of the reactants and products is observed.
  • Question 8
    1 / -0
    Which information can be obtained from Le Chatelier's principle?
    Solution
    Le Chtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to re-establish equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. 

    Hence, option A is correct.
  • Question 9
    1 / -0
    Choose the incorrect statement :
  • Question 10
    1 / -0
    Assertion: Due to common ion effect, the solubility of $$HgI_2$$ is expected to be less in an aqueous solution of KI than in water. But $$HgI_2$$ dissolves in an aqueous solution of KI to form a clear solution.
    Reason: $$I^{\circleddash}$$ ion is highly polarisable.
    Solution
    Due to common ion effect, the solubility of $$HgI_2$$ is expected to be less in an aqueous solution of KI than in water as
    $$HgI_2 + KI \longrightarrow K_2[HgI_4]$$.
    since $$I^{\circleddash}$$ ion is large sized and therefore is highly polarisable.
    But (R) is not the correct explanation of (A)
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