Graph
Recognition Pack

ln[R] vs t plot

Intercept

Set t = 0 in the integrated law.

ln[R] = ln[R]₀ − kt

1. At t = 0 the equation reduces to ln[R] = ln[R]₀.
2. So the intercept on the vertical axis is ln[R]₀, not [R]₀ itself.
3. To recover [R]₀ you would exponentiate: [R]₀ = e^(intercept).
4. The slope, separately, is −k.

Second order reaction

The linear plot

Gap content — the second order integrated law is linear in the reciprocal.

1/[R] − 1/[R]₀ = kt

1. The second order integrated law rearranges to 1/[R] = kt + 1/[R]₀.
2. Comparing with y = mx + c, plotting 1/[R] against t gives a straight rising line.
3. Its slope is +k and its intercept is 1/[R]₀.
4. This form is not derived in NCERT but completes the identification drill.

log(rate) vs log[A] plot

Meaning of the intercept

Taking logs of the rate law separates order into the slope and k into the intercept.

log(Rate) = log k + n·log[A]

1. Comparing with y = mx + c, the slope m is the order n and the intercept c is log k.
2. So the intercept gives log k, and k itself is 10 raised to that value.
3. The order is read from the slope, not the intercept.
4. This single plot therefore yields both unknowns at once.

Two ln[R] vs t lines of different steepness

Interpretation of the steeper line

Slope magnitude is k, and half-life is inversely proportional to k.

slope = −k ; t½ = 0.693/k

1. The magnitude of the slope equals k, so a steeper line means a larger k.
2. A larger k means a faster reaction.
3. Since t½ = 0.693/k, a larger k gives a shorter half-life.
4. Steeper, faster, larger k, shorter half-life — all four go together.

Rate vs concentration straight through the origin

Order

Direct proportionality between rate and concentration defines first order.

Rate = k[R]

1. A straight line through the origin means Rate ∝ [R] exactly.
2. That is the first order rate law, Rate = k[R], with slope k.
3. Zero order would give a horizontal line and second order a parabola.
4. Half order would give a curve that flattens as concentration rises.

First order [R] vs t curve

Why it never reaches zero

An exponential function is positive for every finite argument.

[R] = [R]₀e^(−kt) > 0 for all finite t

1. The concentration follows [R] = [R]₀e^(−kt).
2. The exponential factor becomes very small but never actually reaches zero for finite t.
3. So the curve approaches the time axis without touching it.
4. This is exactly why first order questions ask for 99% or 99.9% completion, never 100%.
5. Contrast with zero order, whose straight line does meet the axis at t = [R]₀/k.

Slope of log([R]₀/[R]) vs t = 0.0301 min⁻¹

k

This plot's slope is k/2.303, so multiply back.

slope = k/2.303 ⟹ k = 2.303 × slope

1. The plotted relation is log([R]₀/[R]) = (k/2.303)t.
2. So k = 2.303 × slope = 2.303 × 0.0301.
3. = 0.0693 min⁻¹.
4. Note this corresponds to t½ = 0.693/0.0693 = 10 min, a tidy result.

Intercept of ln k vs 1/T = 18

A

The intercept of the natural-log form is ln A.

ln k = ln A − Ea/RT ; intercept = ln A

1. The intercept occurs where 1/T = 0, leaving ln k = ln A.
2. So ln A = 18.
3. Exponentiating with base e gives A = e¹⁸.
4. It would be 10¹⁸ only if the vertical axis were log₁₀ k.

An unknown reaction with concentration–time data

The identification procedure

Each order has a unique linearising plot, so testing them in turn identifies the order.

zero → [R] vs t ; first → ln[R] vs t ; second → 1/[R] vs t

1. Each order has its own plot that turns its curve into a straight line.
2. Testing the three plots in turn, whichever produces a straight line identifies the order.
3. The slope of that straight line simultaneously gives k.
4. This is the standard laboratory procedure and the logic behind most graph questions.

[R] vs t plot for zero order

Intercept

Set t = 0 in the integrated law.

[R] = [R]₀ − kt

1. At t = 0 the equation gives [R] = [R]₀.
2. So the intercept on the concentration axis is the initial concentration.
3. The slope, separately, is −k.

Rate vs [A] horizontal at 0.004 mol L⁻¹ s⁻¹

k

A horizontal rate plot means zero order, for which the rate equals k.

Rate = k[A]⁰ = k

1. A horizontal line means the rate is independent of concentration, so the reaction is zero order.
2. For zero order, Rate = k exactly.
3. So k = 0.004 mol L⁻¹ s⁻¹, carrying the same units as the rate.
4. Option (b) gives first order units and is inconsistent with a horizontal plot.

Arrhenius plot with A varied and Ea fixed

Effect on the line

A appears only in the intercept; Ea appears only in the slope.

ln k = (−Ea/R)(1/T) + ln A

1. Comparing with y = mx + c, A occupies the intercept position and Ea the slope position.
2. Changing A therefore alters only the intercept.
3. So the line moves bodily up or down while keeping the same tilt.
4. This clean separation is exactly why the Arrhenius plot is useful — one graph, two independent unknowns.

Slope = −0.025 s⁻¹ on an ln[R] vs t plot

t½

The slope magnitude is k, and first order half-life follows from it.

slope = −k ; t½ = 0.693/k

1. k = −(slope) = 0.025 s⁻¹.
2. t½ = 0.693/0.025.
3. = 27.7 s.

Zero order reaction

Which plot is curved

Only the plot that matches an order's integrated law is straight; the others distort.

[R] = [R]₀ − kt, so ln[R] = ln([R]₀ − kt)

1. [R] vs t is straight for zero order, so (b) is not it.
2. Rate vs t is horizontal, and rate vs [R] is also horizontal — both straight lines.
3. Taking the logarithm of a linear function gives ln([R]₀ − kt), which is not linear in t.
4. So ln[R] vs t is curved for a zero order reaction.

Concentration–time curve

Meaning of instantaneous rate

A derivative is geometrically the slope of a tangent.

r_inst = −d[R]/dt

1. The instantaneous rate is the derivative of concentration with respect to time at that instant.
2. Geometrically, a derivative at a point is the slope of the tangent drawn there.
3. The minus sign only ensures the rate is positive; its magnitude is the slope magnitude.
4. A chord instead of a tangent would give the average rate.

First order reaction, rate plotted against time

Shape

Rate is proportional to concentration, which itself decays exponentially.

Rate = k[R] = k[R]₀e^(−kt)

1. Substituting the integrated law into the rate law gives Rate = k[R]₀e^(−kt).
2. This is an exponential function of time.
3. So the rate versus time plot is an exponential decay curve, with the same shape as the concentration curve.
4. For zero order, by contrast, the rate versus time plot would be horizontal.

Slope of 1/[R] vs t = 0.5 L mol⁻¹ s⁻¹

Order and k

A linear reciprocal plot identifies second order, and its slope is +k.

1/[R] = kt + 1/[R]₀

1. A straight 1/[R] versus t plot is the second order signature.
2. Comparing with y = mx + c, the slope equals +k, positive because 1/[R] rises as [R] falls.
3. So k = 0.5 L mol⁻¹ s⁻¹.
4. The units confirm it: L mol⁻¹ s⁻¹ is exactly a second order rate constant.

Need to distinguish zero from first order

The most direct graph

Half-life behaviour differs sharply and unambiguously between the two orders.

zero: t½ = [R]₀/2k ; first: t½ = 0.693/k

1. For zero order, t½ ∝ [R]₀, giving a rising line through the origin.
2. For first order, t½ is independent of [R]₀, giving a horizontal line.
3. These two shapes are impossible to confuse.
4. Options (b) and (c) concern temperature dependence, which is the same in form for both orders, and (d) is thermodynamic rather than kinetic.

Horizontal ln k vs 1/T plot

Interpretation

Zero slope means zero activation energy.

slope = −Ea/R

1. A horizontal line has zero slope.
2. Since slope = −Ea/R, this requires Ea = 0.
3. With no barrier, the exponential factor equals 1 and k = A, independent of temperature.
4. Order of reaction is unrelated to this plot, so (d) is wrong.

Same k, different [R]₀, zero order

Relation between the lines

The slope carries k and the intercept carries [R]₀.

[R] = [R]₀ − kt

1. The slope is −k for both, and k is the same, so the lines have identical tilt.
2. The intercepts are [R]₀, which differ.
3. Identical slope with different intercepts means parallel lines.
4. They also meet the time axis at different points, [R]₀/k, so the reactions finish at different times.

Rate plotted against time

Meaning of the area under the curve

Integrating a rate over time recovers the total amount reacted.

∫(−d[R]/dt)dt = total change in [R]

1. The vertical axis is a change in concentration per unit time, and the horizontal axis is time.
2. Multiplying them gives a concentration, so the area has units of concentration.
3. Integrating the rate over an interval recovers the total concentration change over that interval.
4. This is the reverse of taking a slope, which recovers a rate from a concentration curve.

First order, [R]₀ doubled

Effect on the ln[R] vs t plot

Initial concentration lives in the intercept; k lives in the slope.

ln[R] = ln[R]₀ − kt

1. The intercept is ln[R]₀, so doubling [R]₀ raises it by ln 2 = 0.693.
2. The slope is −k, which does not depend on [R]₀ at all.
3. So the line shifts up in parallel, keeping the same tilt.
4. This is the graphical statement of the fact that first order half-life is independent of starting concentration.