JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    \[{{\operatorname{N}}_{2}}(g)+3{{\operatorname{H}}_{2}}(g)\rightleftharpoons 2{{\operatorname{NH}}_{3}}(g),{{\operatorname{K}}_{1}}\]                                      [A]                          \[{{\operatorname{N}}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2NO(g),{{\operatorname{K}}_{2}}\]                        [B]                 \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\rightleftharpoons {{H}_{2}}O(g),{{\operatorname{K}}_{3}}\]                          [C]                 The equation for the equilibrium constant of the reaction \[2{{\operatorname{NH}}_{3}}(g)+\frac{5}{2}{{O}_{2}}(g)\rightleftharpoons \]\[2\operatorname{NH}O(g)+3{{\operatorname{H}}_{3}}\operatorname{O}(g),({{K}_{4}})\] in terms of \[{{K}_{1}},{{K}_{2}},\] and \[{{K}_{3}}\]is :     JEE Main Online Paper ( Held On 23  April 2013 )

    A)                  \[\frac{{{\operatorname{K}}_{1}}{{\operatorname{K}}_{2}}}{{{\operatorname{K}}_{3}}}\]                        

    B)                                         \[\frac{{{\operatorname{K}}_{1}}\operatorname{K}_{3}^{2}}{{{\operatorname{K}}_{2}}}\]

    C)                                         \[{{\operatorname{K}}_{1}}{{\operatorname{K}}_{2}}{{\operatorname{K}}_{3}}\]                         

    D)                                         \[\frac{{{\operatorname{K}}_{2}}\operatorname{K}_{3}^{3}}{{{\operatorname{K}}_{1}}}\]

    Correct Answer: D

    Solution :

                      To calculate the-value of\[{{K}_{4}}\] in the given equation we should apply: eqn. (2) + eqn.(3) x 3-eqn. (1) hence \[{{K}_{4}}=\frac{{{K}_{2}}K_{3}^{3}}{{{K}_{1}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner