Q.Prove that for a first order reaction the time required for 99% completion of the reaction is twice the time required for the completion of 90% of the reaction.

For a first-order reaction, only a plot of ln(concentration) versus time is linear; a plot of concentration itself versus time is an exponential curve, not a straight line.

For a first-order reaction, integrating the rate law -d[R]/dt = k[R] gives:

ln[R] = ln[R]0 - kt

This is the equation of a straight line (y = mx + c) when ln[R] is plotted on the y-axis against t on the x-axis, with slope = -k, i.e. k = -slope.

• Option (a), plain concentration [R] vs t, gives a decaying curve (exponential), not a straight line. …

The integrated first-order rate law, k = (2.303/t) log[a/(a−x)], rearranges to log(a−x) = log a − (k/2.303)t, which is linear in t — so a straight line for log(a−x) vs t confirms first-order kinetics.

For a first-order reaction A → products, the rate law is −d[A]/dt = k[A]. Integrating with initial concentration 'a' and amount reacted 'x' at time t (so remaining concentration is a−x):

ln(a−x) = ln a − kt

⇒ log(a−x) = log a − (k/2.303) t

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For a first-order reaction, the integrated rate law $\ln[R] = -kt + \ln[R]_0$ is a straight line when $\ln[R]$ is plotted against time. The slope of this line is $-k$, so the correct answer is (B).

The given logarithmic expression is the standard integrated rate law of a first-order reaction.

For a first-order reaction, rate = -d[R]/dt = k[R]. Integrating between limits [R]0 (at t=0) and [R] (at time t):

ln([R]0/[R]) = kt, or in base-10 form, k = (2.303/t) log10([R]0/[R]).

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For a first order reaction, plotting ln[R] against time gives a straight line with a negative slope equal to -k, because the integrated rate law is linear in ln[R].

For a first order reaction, Rate = k[R]

Integrating: d[R]/[R] = -k dt

On integration: ln[R] = ln[R]0 - kt

This equation has the form y = c + mx, where y = ln[R], x = t, intercept c = ln[R]0, and slope m = -k.

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For a first-order reaction, the plot of $\ln[R]$ versus time is a straight line with slope $= -k$, where $k$ is the rate constant. The slope is negative and its magnitude gives the rate constant directly.

First-order gas evolution: k = (2.303/t)·log[V∞/(V∞ − Vt)]; the printed numbers (10 L at 10 min, 100 L total) give k = (2.303/10)·log(100/90), which is not any of the four options — likely a misprint.

Because the half-life is independent of concentration, the reaction is first order. For a gas-evolution first-order reaction the volume of N₂ formed measures the extent of reaction: V∞ ∝ initial reactant a, and (V∞ − Vt) ∝ remaining reactant (a − x). Hence

k = (2.303/t)·log[V∞/(V∞ − Vt)].

Substituting V∞ = 100 L, Vt = 10 L, t = 10 min:

k = (2.303/10)·log(100/90) = (2.303/10)·log(1.11) ≈ 0.0046 min⁻¹.

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The integrated first-order rate law gives K = (2.303/t) log10 [a/(a-x)], so option (a) is correct.

For a first-order reaction, rate = -d[A]/dt = k[A]. On integrating between the initial concentration a (at t = 0) and (a-x) after time t (where x has reacted):

k = (1/t) ln [a/(a-x)]

Converting the natural logarithm to base 10 (ln = 2.303 log10):

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