11th Class Chemistry Equilibrium Question Bank 11th CBSE Chemistry Ionic Equilibrium

  • question_answer For the reaction \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g),\]at \[400\,K,\,{{K}_{p}}=41\]. Find the value of \[{{K}_{p}}\] for each of the following reactions at the same temperature: (i) \[2N{{H}_{3}}(g)\,\rightleftharpoons {{N}_{2}}(g)+3{{H}_{2}}(g)\] (ii) \[\frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g)\rightleftharpoons N{{H}_{3}}(g)\] (iii)\[2{{N}_{2}}(g)+6{{H}_{2}}(g)\rightleftharpoons 4N{{H}_{3}}(g)\]

    Answer:

                    (i) It is the reverse of the given reaction. Hence, \[{{K}_{p}}=\frac{1}{41}\] (ii) It is obtained by dividing the given equation by 2. Hence, \[{{K}_{p}}=\sqrt{41}\] (iii) It is obtained by multiplying the given equation by 2. Hence, \[{{K}_{p}}={{(41)}^{2}}\].


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