If we plot graph for the first order reaction involving reactant R, then it will be a straight line if plotted for: |
\(\frac{[R]}{[R]_0}\) against time, t \([R]_0\) against time t \(log\frac{[R]}{[R]_0}\) against time, t \([R]\) against time, t |
\(log\frac{[R]}{[R]_0}\) against time, t |
The correct answer is option 3.\(log\frac{[R]}{[R]_0}\) against time, t For a first-order reaction, the rate law is given by: \(-\frac{d[R]}{dt} = k[R] \) Where: \([R]\) is the concentration of the reactant at time \(t\), \([R]_0\) is the initial concentration of the reactant, \(k\) is the rate constant. The integrated form of the first-order rate law is: \(ln\frac{[R]}{[R]_0} = -kt \) This equation indicates that a plot of \(ln\frac{[R]}{[R]_0}\) against time \(t\) for a first-order reaction will yield a straight line. Therefore, the correct answer is: 3. \(log\frac{[R]}{[R]_0}\) against time, \(t\) |