\(\dfrac{13}{20}\)
\(\dfrac{1}{20}\)
\(\dfrac{1}{4}\)
\(\dfrac{1}{5}\)
\(\dfrac{3}{5}\)
Given Data
\( P(E)=\dfrac{1}{4},P(E∪F)=\dfrac{2}{5}\)
We know that,
\(P(E ∪ F) = P(E) + P(F) - P(E ∩ F)\)
\(P(E ∩ F) = \dfrac{1}{4} + a - \dfrac{2}{5}\)---------(1)
Now, \(P(E ∩ F) = P(E) × P(F)\)
\(P(E ∩ F) = (1/4) × a\)----------(2)
Now, we can equate (1) and (2) to get the value of \('a’\)
\(\dfrac{1}{4} × a = \dfrac{1}{4} + a - \dfrac{2}{5}\)
\(⇒\dfrac{1}{4} × a -a = \dfrac{-3}{20}\)
\(⇒\dfrac{-3a}{4}=\dfrac{-3}{20}\)
\(⇒a=\dfrac{1}{5}\) (_Ans)
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: