A)
Rate in terms of \[{{N}_{2}}\] \[(mol\,{{L}^{-1}}se{{c}^{-1}})\]
Rate in terms of \[{{H}_{2}}\] \[(mol\,{{L}^{-1}}se{{c}^{-1}})\]
\[2\times {{10}^{-4}}\]
\[2\times {{10}^{-4}}\]
B)
Rate in terms of \[{{N}_{2}}\] \[(mol\,{{L}^{-1}}se{{c}^{-1}})\]
Rate in terms of \[{{H}_{2}}\] \[(mol\,{{L}^{-1}}se{{c}^{-1}})\]
\[3\times {{10}^{-4}}\]
\[1\times {{10}^{-4}}\]
C)
Rate in terms of \[{{N}_{2}}\] \[(mol\,{{L}^{-1}}se{{c}^{-1}})\]
Rate in terms of \[{{H}_{2}}\] \[(mol\,{{L}^{-1}}se{{c}^{-1}})\]
\[1\times {{10}^{-4}}\]
\[3\times {{10}^{-4}}\]
D)
Rate in terms of \[{{N}_{2}}\] \[(mol\,{{L}^{-1}}se{{c}^{-1}})\]
Rate in terms of \[{{H}_{2}}\] \[(mol\,{{L}^{-1}}se{{c}^{-1}})\]
\[2\times {{10}^{-1}}\]
\[2\times {{10}^{-3}}\]
Correct Answer: C
Solution :
[c]\[{{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}\]; Rate is given by any of the expressions \[-\frac{d\left[ {{N}_{2}} \right]}{dt}=-\frac{1d\left[ {{H}_{2}} \right]}{3dt}\text{=}\frac{1}{2}\frac{d\left( N{{H}_{3}} \right]}{dt}\] Rate of disappearance of \[{{N}_{2}}=\frac{1}{2}\]the rate of formation of \[N{{H}_{3}}=1\times {{10}^{-4}}\] Rate of disappearance of \[{{H}_{2}}=3/2\] the rate of formation of \[N{{H}_{3}}=3\times {{10}^{-4}}\]
You need to login to perform this action.
You will be redirected in
3 sec
