CUET UG Chemistry Booster Test – 3 Integrated Rate Equations

The differential rate equation expresses the rate of reaction in terms of the instantaneous change in concentration with time. Experimentally, determining the instantaneous rate requires drawing tangents to concentration-time curves and calculating their slopes at different points. This process is inconvenient and often inaccurate. To overcome this difficulty, the differential rate equation is integrated. Integration produces an equation that directly relates concentration, time and the rate constant. Such integrated rate equations allow experimental data to be analyzed more conveniently and accurately without repeatedly determining tangent slopes. Therefore, the primary reason for integration is to replace cumbersome instantaneous rate measurements with a direct mathematical relationship between concentration and time.

For the reaction: A(g) → B(g) + C(g) Let the decrease in pressure of A be x. Then: pₜ = pᵢ + x Hence: x = pₜ − pᵢ The partial pressure of A at time t is: pA = pᵢ − x = pᵢ − (pₜ − pᵢ) = 2pᵢ − pₜ Substituting into the first order integrated equation: k = (2.303/t) log(pᵢ/pA) gives: k = (2.303/t) log[pᵢ/(2pᵢ − pₜ)] Therefore, Option A is correct.

Zero order reactions occur when the reaction rate becomes independent of reactant concentration. This situation commonly arises when all active sites on a catalyst surface are occupied or when enzymes become saturated with substrate molecules. Examples given in NCERT include enzyme-catalysed reactions and reactions occurring on metal surfaces such as the decomposition of ammonia on platinum and HI on gold. Since the rate remains constant after saturation, these reactions exhibit zero order kinetics. Therefore, Option C is correct.

At high pressure, a large number of ammonia molecules adsorb on the platinum surface. Eventually, all available catalytic sites become occupied, producing a saturated surface. Once saturation occurs, increasing ammonia concentration further cannot increase the number of adsorbed molecules. As a result, the reaction rate remains constant and no longer depends on reactant concentration. This is the defining characteristic of a zero order reaction. Therefore, the decomposition of ammonia on platinum exhibits zero order kinetics under such conditions.

The integrated zero order equation is: [R] = [R]₀ − kt For complete consumption of the reactant: [R] = 0 Substituting: 0 = [R]₀ − kt kt = [R]₀ Therefore: t = [R]₀/k This expression gives the total time required for complete disappearance of the reactant in a zero order reaction. Hence, the correct answer is A.

For a zero order reaction: [R] = [R]₀ − kt Comparing with the straight-line equation: y = mx + c Slope = -k The unit of concentration [R] is mol L⁻¹ and the unit of time is s. Therefore: Unit of slope = (mol L⁻¹)/s = mol L⁻¹ s⁻¹ Since slope equals -k, the rate constant of a zero order reaction also has the same unit. Thus, the correct answer is C. mol L⁻¹ s⁻¹.

From the passage: pₜ = pᵢ + x Therefore: x = pₜ − pᵢ The partial pressure of A at time t is: pA = pᵢ − x Substituting x: pA = pᵢ − (pₜ − pᵢ) pA = 2pᵢ − pₜ This expression is used in deriving the integrated rate equation for first order gas phase reactions. Thus, the correct answer is B. 2pᵢ − pₜ.

For a zero order reaction: [R] = [R]₀ − kt Similarly, if pressure is used instead of concentration: pA = pA₀ − kt Comparing with: y = mx + c Slope = -k Thus, a graph of pA versus t gives a straight line whose slope is equal to the negative of the rate constant. Therefore, the correct answer is B. The negative rate constant (-k).

For a first order reaction: k = (2.303/t) log([R]₀/[R]) As the percentage completion increases, the amount of reactant remaining decreases. Therefore, greater completion requires a longer reaction time. Given: 1. 50% completion 2. 99.9% completion 3. 90% completion 4. 99% completion Arranging in decreasing order of time required: 1. 99.9% completion → Maximum time 2. 99% completion 3. 90% completion 4. 50% completion → Minimum time Hence, the decreasing order is: 2 > 4 > 3 > 1 Thus, the correct answer is A. 2, 4, 3, 1.

Natural radioactive decay processes follow first order kinetics because the rate depends on the number of undecayed nuclei present. Hydrogenation of ethene is a standard NCERT example of a first order reaction. The decomposition of N₂O₅ is also a classic first order reaction. However, the decomposition of HI on a gold surface is a zero order reaction under conditions where the catalyst surface is saturated. Therefore, Statements 1, 2 and 3 are correct, while Statement 4 is incorrect. Thus, the correct answer is A. 1, 2 and 3 are correct.

For a first order reaction: ln[R]₁ = -kt₁ + ln[R]₀ ln[R]₂ = -kt₂ + ln[R]₀ Subtracting the first equation from the second: ln[R]₂ - ln[R]₁ = -k(t₂ - t₁) Using logarithmic properties: ln([R]₂/[R]₁) = -k(t₂ - t₁) Multiplying by -1: ln([R]₁/[R]₂) = k(t₂ - t₁) Therefore: k = (1/(t₂ - t₁)) ln([R]₁/[R]₂) Hence, Option A is correct.

The integrated first order equation is: [R] = [R]₀e⁻ᵏᵗ Dividing by [R]₀: [R]/[R]₀ = e⁻ᵏᵗ The term e⁻ᵏᵗ therefore represents the fraction of reactant remaining after time t. As time increases, the value of e⁻ᵏᵗ decreases exponentially, showing the gradual consumption of reactant. Thus, the correct answer is A. e⁻ᵏᵗ.

For first order kinetics: [R] = [R]₀e⁻ᵏᵗ Hence, the [R] versus t graph is an exponential decay curve. Taking logarithms: ln[R] = -kt + ln[R]₀ Thus, ln[R] versus t gives a straight line with slope -k and intercept ln[R]₀. Also: log([R]₀/[R]) = (k/2.303)t This graph is a straight line passing through the origin. Therefore: 1 → b 2 → d 3 → a 4 → c Hence, Option C is correct.

For a first order reaction: ln[R] = -kt + ln[R]₀ or log[R] = -(k/2.303)t + log[R]₀ This equation represents a straight line with a negative slope. Therefore, when a graph of log[R] versus t is linear and decreasing, it is a characteristic signature of first order kinetics. Hence, the correct answer is B. First order.

For 99.9% completion: [R] = 0.001[R]₀ Using: k = (2.303/t)log([R]₀/[R]) k = (2.303/t)log(1000) Since: log(1000) = 3 t₉₉.₉% = 6.909/k For first order reactions: t₁/₂ = 0.693/k Therefore: t₉₉.₉%/t₁/₂ = (6.909/k)/(0.693/k) ≈ 9.97 ≈ 10 Hence, the correct answer is B. Approximately 10.

For a first order reaction: t₁/₂ = 0.693/k Given: k = 5.5 × 10⁻¹⁴ s⁻¹ Substituting: t₁/₂ = 0.693/(5.5 × 10⁻¹⁴) = 0.126 × 10¹⁴ = 1.26 × 10¹³ s Therefore, the half-life of the reaction is: 1.26 × 10¹³ s Hence, the correct answer is D.

For a first order reaction: t₁/₂ = 0.693/k Since the half-life depends only on the rate constant, every reduction to half the current concentration requires the same amount of time. Thus: 100% → 50% takes one half-life. 50% → 25% also takes one half-life. 25% → 12.5% again takes one half-life. This unique property is characteristic of first order kinetics. Therefore, Option A is correct.

Half-life is defined as the time required for the concentration of a reactant to decrease to one-half of its initial value. Since half-life measures time, its unit must always be a unit of time such as: • seconds (s) • minutes (min) • hours (h) • days (d) This remains true for zero order, first order and all other reaction orders. Therefore, the correct answer is C.

For a zero order reaction: t₁/₂ = [R]₀/(2k) This shows: t₁/₂ ∝ [R]₀ If the new initial concentration becomes: [R]₀' = [R]₀/4 Then: t₁/₂' = ([R]₀/4)/(2k) = (1/4)([R]₀/(2k)) = (1/4)t₁/₂ Therefore, the new half-life becomes one-fourth of the original value. Thus, the correct answer is C.

For a first order reaction: k = (1/t) ln([R]₀/[R]) At half-life: [R] = [R]₀/2 Substituting: k = (1/t₁/₂) ln([R]₀/([R]₀/2)) = (1/t₁/₂) ln 2 Therefore: t₁/₂ = (ln 2)/k Since: ln 2 = 0.693 the equation becomes: t₁/₂ = 0.693/k Hence, the value 0.693 originates from ln 2. Therefore, the correct answer is B.