A) \[n-m\]
B) \[{{2}^{n-m}}\]
C) \[\frac{1}{{{2}^{m+n}}}\]
D) \[m+n\]
Correct Answer: B
Solution :
[b] \[\text{Rat}{{\text{e}}_{1}}=K{{[A]}^{n}}{{[B]}^{m}};\,\,\text{Rat}{{\text{e}}_{2}}=K{{[2A]}^{n}}{{\left[ \frac{1}{2}B \right]}^{m}}\]\[\therefore \frac{Rat{{e}_{2}}}{Rat{{e}_{1}}}=\frac{K{{[2A]}^{n}}{{\left[ \frac{1}{2}B \right]}^{m}}}{K{{[A]}^{n}}{{[B]}^{m}}}\] \[={{(2)}^{n}}{{\left( \frac{1}{2} \right)}^{m}}={{2}^{n}}\cdot {{2}^{-m}}={{2}^{n-m}}\]
You need to login to perform this action.
You will be redirected in
3 sec
