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Equilibrium Test - 21

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Equilibrium Test - 21
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  • Question 1
    1 / -0
    An endothermic reaction which proceeds with a decrease in volume will give a maximum yield of the products at...
    Solution
    When the pressure on the system is increased, the volume decreases proportionately. The total number of moles per unit volume increases. According to Le-Chatelier's principle, the equilibrium shift in the direction in which there is decrease in number of moles. If there is no change in number of moles of gases in a reaction, a pressure change does not affect the equilibrium.
     If the temperature at equilibrium is increased reaction will proceed in the direction in which heat can be used. Thus increase in temperature will favour the forward reaction for endothermic reaction. 
    Similarly, increase in temperature will favour the backward reaction in exothermic reactions.
  • Question 2
    1 / -0
    In the reaction $$A + B \rightleftharpoons AB$$, if the concentration of A and B is increased by a factor of 2, it will cause the equilibrium concentration of AB to change to :
    Solution
    For $$A+B\rightleftharpoons AB$$
    $$K=\frac {[AB]}{[A][B]}$$
    When $$2A+2B\rightleftharpoons 2AB$$
    $$K=2AB$$
  • Question 3
    1 / -0
    A tenfold increase in pressure on the reaction $$N_2(g) + 3H_2(g)\rightleftharpoons 2NH_3(g)$$ at equilibrium constant $$K_p$$ will :
    Solution
    As we know,
    $$K_P$$ depends on temperature only so by increasing pressure, eqm constant doesnt even change and it will remain same.
  • Question 4
    1 / -0
    Low temperature is favourable for .......... reactions.
    Solution
     If the temperature at equilibrium is increased reaction will proceed in the direction in which heat can be used. Thus increase in temperature will favour the forward reaction for endothermic reaction.
    Similarly, increase in temperature will favour the backward reaction in exothermic reactions.
  • Question 5
    1 / -0
    If the temperature of the system at equilibrium is increased:
  • Question 6
    1 / -0
    The reaction $$CaCO_3(g)\rightleftharpoons CaO(g)+CO_2(g)$$ proceeds to completion if ....
    Solution
    The reaction $$CaCO_3(g)\rightleftharpoons CaO(g)+CO_2(g)$$ proceeds to completion if $$CO_2$$ escapes.
    As carbon dioxide escapes, its concentration decreases. So the equilibrium will shift in the forward direction so that more and more carbon dioxide is formed to nullify the effect of decrease in the carbon dioxide concentration.
  • Question 7
    1 / -0
    When a product is removed from the system the equilibrium is reaction favored.
    Solution
    When a product is removed from the system which is at equilibrium forward reaction is favored.
    For example, the reaction $$CaCO_3(g)\rightleftharpoons CaO(g)+CO_2(g)$$ proceeds to completion if $$CO_2$$ escapes.
    As carbon dioxide escapes, its concentration decreases. So the equilibrium will shift in the forward direction so that more and more carbon dioxide is formed to nullify the effect of decrease in the carbon dioxide concentration.
  • Question 8
    1 / -0
    The degree of dissociation of $$PCl_5$$ will be more at .....
    Solution
    The degree of dissociation of $$PCl_5$$, will be more at low pressure.
    $$PCl_5 (g) \rightleftharpoons PCl_3 (g) +Cl_2 (g)$$
    $$\Delta n = 1+1-1=-1$$
    The reaction proceeds with decrease in the number of moles of gaseous species.
    Hence, at low pressure, the equilibrium will shift in the forward direction so that more and more of $$PCl_5$$ is dissociated. This will nullify the effect of low pressure.
    This is according to Le chatelier's principle.
  • Question 9
    1 / -0
    When the system $$2HI(g)\rightleftharpoons H_2(g) + I_2(g)$$ is at equilibrium. if an inert gas is introduced, then dissociation of HI is will 
    Solution
    When the system $$2HI(g)\rightleftharpoons H_2(g) + I_2(g)$$ is at equilibrium. if inert gas is introduced,then dissociation of HI is will not change . For the above reaction, $$\Delta n = 1+1-2=0$$. There is no change in the number of moles of gaseous species. Hence, the change in pressure has no effect on the equilibrium.
  • Question 10
    1 / -0
    The formation of ammonia by Haber's process is favored by high pressure.
    Solution
    The formation of ammonia by Haber's process is. favored by high pressure.
    $$N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g)$$
    $$\Delta n = 2- (1+3) = 2-4 = -2 $$
    Since the reaction occurs with decrease in the number of moles of gaseous species, high pressure will shift the equilibrium towards right so as to nullify the effect of increased pressure. This is as per Le chatelier's principle.
    Thus more ammonia will b formed.
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