The figure shows the plot of log([R]0/[R]) against time for a first o…

The correct answer is Option (2) → 0, because at t = 0, [R] = [R]₀ and log 1 = 0

• Value read from the figure — The line starts at the origin, so at t = 0 the value of log([R]₀/[R]) is 0.

• Reason — At the start none of the reactant has been used, so [R] = [R]₀; the ratio [R]₀/[R] is 1 and log 1 = 0. The line follows the equation below, of the form y = mx with slope k/2.303 and no intercept.

      \( \displaystyle \log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} = \frac{kt}{2.303} \)

Why the other options are incorrect

Option (1) 1, because at t = 0, [R] = [R]₀ — Incorrect. The ratio [R]₀/[R] is 1, but the plotted quantity is its logarithm, log 1 = 0; the line starts at 0, not at 1.

Option (3) log [R]₀, because the line starts from the initial concentration — Incorrect. An intercept related to [R]₀ belongs to the plot of ln[R] against t; this line has no intercept and starts at the origin.

Option (4) 0, because at t = 0 the rate constant k is zero — Incorrect. The value is 0, but k does not change with time; the line has the same slope, k/2.303, all along its length.

NCERT (2026-27 edition):
3.3.2 First Order Reactions – “The first order rate equation (3.10) can also be written in the form”

    \( \displaystyle k = \frac{2.303}{t}\log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} \qquad (3.15) \)

    \( \displaystyle \log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} = \frac{kt}{2.303} \)

“If we plot a graph between log [R]₀/[R] vs t, (Fig. 3.5), the slope = k/2.303”

Fig. 3.5: Plot of log [R]₀/[R] vs time for a first order reaction