Chemical Kinetics

In a certain gaseous reaction between X and Y ; X + 3Y → XY₃

The initial rates are reported as follows –

The rate law is -

\frac{k_{35^\circ}}{k_{34^\circ}}>1, this means that –

Reaction : 2Br^- + H₂O₂ + 2H⁺ → Br₂ + 2H₂O take place in two steps :

(i) $\mathrm{B}{\mathrm{r}}^{-}+{\mathrm{H}}^{+}+{\mathrm{H}}_2{\mathrm{O}}_2\stackrel{\mathrm{slow}}{\to }\mathrm{HOBr}+{\mathrm{H}}_2\mathrm{O}$

(ii) $\mathrm{HOBr}+\mathrm{B}{\mathrm{r}}^{-}+{\mathrm{H}}^{+}\stackrel{\mathrm{fast}}{\to }{\mathrm{H}}_2\mathrm{O}+\mathrm{B}{\mathrm{r}}_2$

The order of the reaction is —

For the reaction NO₂ + CO → CO₂ + NO the experimental rate expression is

− dc/dt = k[NO₂]². The number of molecules of CO involved in the slowest step will be —

For the reaction A → B, data of initial concentration and corresponding half life period are given in the tabular form –

[A] 1M 2M 4M

T_1/2 300s 600s 1200s

The order of the reaction is –

For the reaction 2H₂ + 2NO → N₂ + 2H₂O

the following mechanism has been proposed-

(i) 2NO ⇌ N₂O₂ (fast)

(ii) N₂O₂ + H₂ → N₂O + H₂O (r.d.s)

(iii) N₂O + H₂ → N₂ + H₂O (fast)

then what will be the rate law of this reaction?

For consecutive reaction $\mathrm{A}\stackrel{k_1}{\to }\mathrm{B}\stackrel{k_2}{\to }\mathrm{C}\stackrel{k_3}{\to }\mathrm{D}$

k₁ << k₂ << k₃. order of reaction will be

What is the order of a reaction whose rate law is $r=K{C}_A^{3/2}{C}_B^{-1/2}$

The instantaneous rate of disappearance of the MnO₄^- ion in the following reaction is 4.56 × 10⁻³ Ms⁻¹. 2MnO₄^- + 10 I^- + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O The rate of appearance of I₂ is -

For the decomposition of N₂O₅ at a particular temperature, according to the equation :

2N₂O₅ → 4NO₂ + O₂

N₂O₅ → 2NO₂ + 1/2 O₂

the activation energy are E₁ & E₂ respectively, then

Which curve represents zero order reaction ?

For Ist order reaction 2x(g) → 3y(g) + 2z(g)

Assuming ideal nature of gas, time taken for 75% completion of reaction will be

A following mechanism has been proposed for a reaction -

2A + B → D + E

A + B → C + D (slow)

A + C → E (fast)

The rate law expression for the reaction is -

What is the order of a chemical reaction A + 2B →C, if the rate of formation of 'C' increases by a factor of 2.82 on doubling the concentration of 'A' and increases by a factor of 9 on tripling the concentration of 'B'–

A hypothetical reaction A₂ + B₂ → 2AB follows the mechanism as given below -

A₂ ⇌ A + A (fast)

A + B₂ → AB + B (slow)

A + B → AB (fast)

The order of the over all reaction is -

For the reaction ; 2HI → H₂ + I₂, the expression

$-\frac{1}{2}\frac{d[\mathrm{HI}]}{dt}$ . represents -

Which of the following graph is correct for the reaction :

A + B → C + D + 100 K cal -

For a reaction 3A → products, it is found that the rate of reaction doubles, if concentration of A is increased four times, calculate order of reaction

For a second order reaction, the graph plotted between [A]⁻¹ and time 't' is shown below - θ = tan⁻¹ (0.5) and OP = 2L mol⁻¹. Thus rate of the reaction at start is –

Consider the plots given below, for the type of reaction nA ⟶ B + C

These plots respectively corresponds to the reaction order

For a reaction AB₅ → AB + 4B The rate can be expressed in following ways $-\frac{d[\mathrm{A}{\mathrm{B}}_5]}{dt}=K[\mathrm{A}{\mathrm{B}}_5];\frac{d[\mathrm{B}]}{dt}=K_1[\mathrm{A}{\mathrm{B}}_5]$ So the correct relation between K and K₁ is-

The rate of certain hypothetical reaction A + B + C → products, is given, by $r=-\frac{dA}{dt}=K[A{]}^{1/2}[B{]}^{1/3}[C{]}^{1/4}$ The order of reaction is given by -

Consider a reaction A → B + C. If the initial concentration of A was reduced from 4 M to 2 M in 1 hour and from 2 M to 1 M in 0.5 hour, the order of the reaction is-

Figure depicts the change in conc. of species x & y for reaction 2x → y, as a function of time the point of inter section of two curves represents

The progress of the reaction A ⇌ nB, with time is represented by the graph given below- The value of n is

The following data pertain to reaction between A and B :

Which of the following conclusions can be drawn from the above data —

(a) Rate constant of the reaction 10^-4

(b) Rate law of the reaction is k[A][B]

(c) Rate of reaction increases four times on doubling the concentration of both the reactant,

Select the correct answer – codes :–

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C ? (R = 8.314 J mol^–1 K^–1)

A reaction follows the given concentration (C) vs time graph. The rate for this reaction at 20 seconds will be –

A reaction takes place in three steps. The rate constants are k₁, k₂ and k₃. The over all rate constant $k=\frac{k_1{k}_3}{k_2}$ . If (energy of activation) E₁, E₂ and E₃ are 60, 30 and 10 kJ. The overall energy of activation is –

A reaction rate constant is given by : $k=1.2\times 10^{14}{e}^{-25000/(RT)}{\sec }^{-1}$. It means –

In the reaction : A + 2B + C → D + 2E

The rate of reaction remains unchanged if the conc. of B is doubled and that of A and C is kept constant. What is the order with respect to B —

The chemical reaction 2O₃ → 3O₂ proceeds as follows -

O₃ ⇌ O₂ + O ............(fast)

O + O₃ → 2O₂ ............(slow)

The rate law expression should be -

Select the rate law that corresponds to the data shown for the reaction A + B → C

The t_1/2 for the first order reaction,

PCl₅(g) ⟶ PCl₃(g) + Cl₂(g) is 20 min. The time in which the conc. of PCl₅ reduces to 25% of the initial conc. is close to-

t_1/2 = constant, confirms the order of the reaction as one. a²t_1/2 = constant confirms that the reaction is of –

In the formation of sulphur trioxide by the contact process,

2SO₂ (g) + O₂ (g) ⟶ 2SO₃ (g)

The rate of reaction is expressed as − d(O₂)/dt = 2.5 × 10⁻⁴ mol L⁻¹ sec⁻¹

The rate of disappearance of (SO₂) will be -

For a hypothetical reaction A → L, the rate expression is-

$\text{rate}=-\frac{d{C}_A}{dt}$

A reaction takes place in various steps. The rate const. for 1st, 2nd, 3rd & 5th steps are k₁,k₂, k₃ & k₅ resp. The overall rate constant is given by $k=\frac{k_2}{k_3}{\left(\frac{k_1}{k_5}\right)}^{1/2}$. If activation energy are 40, 60, 50 & 10 kj/mole resp. The overall energy of activation (kJ/mole) is -

For a reaction 2P + Q → S, following data were collected. Calculate the overall order of the reaction.

The reaction L → M is started with 10 gm of L. After 30 & 90 min. 5gm & 1.25 gm of L are left respectively. The order of reaction is

Using the data given below the order and rate constant for the reaction :

CH₃CHO(g) → CH₄(g) + CO(g)

In the following reaction : x A→yB

$\log \left[-\frac{d[A]}{dt}\right]=\log \left[\frac{d[B]}{dt}\right]+0.3$

where −ve sign indicates rate of disappearance of the reactant. Thus, x : y is –

For a gaseous reaction, following data is given

$\mathrm{A}\to \mathrm{B},k_1=10^{15}{e}^{\frac{-2000}{T}}\text{and}\mathrm{C}\to \mathrm{D},k_2=10^{14}{e}^{\frac{-1000}{T}}$

the temperature at which k₁ = k₂ is

In the following reaction A →B + C, rate constant is 0.001 M/sec. If we start with 1M of A the conc. of A & B after 10 min are respectively

For a reaction, the rate constant is expressed as, k = A.e^(-40000/T)

The energy of the activation is -

A reaction A₂ + B₂ → 2AB occurs by the following mechanism -

A₂ → A + A ........(slow)

A + B₂ → AB + B ......(fast)

A + B → AB ........(fast) its order would be -

Consider the data given below for hypothetical reaction A → X.

From the above data, the order of reaction is-

A first order reaction takes 69.3 minute  for 50% completion. How much time will be needed for 80% completion -

In acidic medium, the rate of reaction between (BrO₃)^- and Br^- ion is given by the expression : − $-\frac{d(\mathrm{Br}{\mathrm{O}}_3^{-})}{dt}=k[\mathrm{Br}{\mathrm{O}}_3^{-}][\mathrm{B}{\mathrm{r}}^{-}][{\mathrm{H}}^{+}{]}^2$ it means -