Chemical Kinetics

Given-

Initial concentration(R₁) = 0.03M

Final concentration(R₂) = 0.02M

Time = 25 mins.

The formula for average rate of the reaction is,

→ Equation 1

∵ {R} = (R₂)-( R₁), the equation 1 is written as,

= 4 × 10^-4 mol L^-1 min^-1

The average rate of reaction in seconds is given by,

=

(dividing by 60 to convert minutes to seconds)

= 6.6 × 10^-6 mol L^-1 s^-1

The average rate of the reaction in minutes is4 × 10^-4 mol L^-1 min^-1 and in seconds is 6.6 × 10^-6 mol L^-1 s^-1

Given-

Initial concentration(A₁) = 0.5M

Final concentration(A₂) = 0.4M

Time = 10 mins.

The formula for average rate of the reaction is,

→ Equation 1

∵ {A} = (A₂)-( A₁), the equation 1 is written as,

=

= 0.005 mol L^-1 min^-1

= 5 × 10^-3 mol L^-1 min^-1

The average rate of the reaction is5 × 10^-3 mol L^-1 min^-1

The order of the reaction is sum of the powers of concentration of reactants in the rate law. According to this,

The order of the reaction = 1/2 + 2

=

The order of the reaction is 2.5

Let the reaction be X→Y

As, this reaction follows second order kinetics, the rate of the reaction will be,

Rate = k(X)² → Equation 1

Since the concentration of X is increased three times, Equation 1

will become,

Rate = k(3X)²

= k × 9(X)²

∴ The rate of formation of Y will become 9 times.

Thus, the rate of formation of Y will become 9 times

Given-

Rate constant, k = 1.15 × 10^-3 s^-1

Initial quantity, R₀ = 5 g

Final quantity, R = 3 g

According to the formula of first order reaction,

t =

t = 443.8 s

t = 4.438 × 10² s

The time taken for 5g of the reactant to reduce to 3 g is 4.438 × 10² s

Given-

60 mins

Using the formula for half life, , we get,

∴ k = 1.155 × 10^-2 min^-1

The rate constant of the reaction, k is1.155 × 10^-2 min^-1

The rate constant of a chemical reaction normally increases with increase in temperature. It is observed that for a chemical reaction with rise in temperature by 10°C, the rate constant is nearly doubled and this temperature dependence of the rate of a chemical reaction can

be accurately explained by Arrhenius equation,

k = A

Thus, the rate constant of a chemical reaction normally increases with increase in temperature.

Given-

Initial temperature, T₁ = 298 K

Final temperature, T₂ = 298 K + 10 K = 308 K

Knowing that the rate constant of a chemical reaction normally increases with increase in temperature, we assume that,

Initial value of rate constant, k₁ = k

Final value of rate constant, k₂ = 2k

Using Arrhenius equation,

→ Equation 1

where, R = 8.314 J K^-1 mol^-1 (gas constant).

Substituting all the values in equation 1, we get,

log 2 =

E_a =

E_a =

E_a = 52897 J mol^-1

E_a = 52.897 kJ mol^-1

The energy of activation, E_a is 52.897 kJ mol^-1

Given-

Energy of activation, E_a= 209.5 kJ mol^–1

E_a in joules = 209.5 × 1000 = 209500 J mol^–1

Temperature, T = 581K

Gas constant, R = 8.314 J K^-1 mol^-1

Using Arrhenius equation,

k = A

In this equation, the term represents the number of molecules which have kinetic energy greater than the activation energy, E_a.

∴ The number of molecules = → Equation 1

Substituting all the known values in equation 1, we get,

⇒

= e^-43.3708

Finding the value of anti ln(43.3708), we get, 1.47 × 10^-19

The fraction of molecules of reactants having energy equal to or greater than activation energy is1.47 × 10^-19

(i) 3NO(g) → N₂O(g)

Rate = k[NO]²

Here rate law is dependent on concentration of NO only.And power raised to concentration of NO is 2. Hence order of reaction is 2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in . So, unit of K will be

i.e. [concentration]^-1[ time]^-1 . If concentration is in mol L^-1 and time is in sec. then dimensions of k will be : mol^-1 L sec^-1

(ii) H₂O₂(aq) + 3I^-(aq) + 2H⁺ → 2 H₂O(l) + I₃^-

Rate = k[H₂O₂][I^-]

Here rate law is dependent on concentration of H₂O₂ and I^-. And power raised to concentration of H₂O₂ is 1, and power raised to concentration of I^- is 1. order is sum of powers hence, order of given reaction is 2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]² i.e. [concentration]^-1[ time]^-1. If concentration is in mol L^-1 and time is in sec then dimensions of k will bemol^-1 L sec^-1

(iii) CH₃CHO(g) → CH₄(g) + CO(g)

Rate = k[CH₃CHO]^3/2

Here rate law is dependent on concentration of CH₃CHO.And power raised to CH₃CHO is 3/2. Hence order of reaction is 3/2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]^3/2 i.e. [concentration]^-1/2[time]^-1.

If concentration is in mol L^-1 and time is in sec. Then dimensions of K will be mol^-1/2L^1/2 sec^-1

(iv) C₂H₅Cl (g) → C₂H₄(g) + HCl(g)

Rate = k[C₂H₅Cl]

Here rate law is dependent on concentration of C₂H₅Cl. And power raised to C₂H₅Cl is 1. Hence order of reaction is 1.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]¹. i.e. [time]^-1

If concentration is in mol L^-1 and time is in sec. Then dimensions of K will be sec^-1.

a) Rate = k[A][B]², Rate = 2.0 × 10^-6[0.1][0.2]²

Rate = 8 × 10^-9 mol L^-1sec^-1

b)

Situation when A is remained 0.06 mol L^-1

Now, According to rate law, Rate = k[A][B]²

Rate = 2 × 10^-6[0.06][0.18]²

i.e. Rate = 3.888 × 10^-9 mol L^-1sec^-1

Initial rate of reaction is 8 × 10^-9 mol L^-1sec^-1. and rate when concentration of A is 0.06 mol L^-1 is 3.888 × 10^-9 mol L^-1sec^-1.

2NH₃(g) → N₂(g) + 3H₂(g)

Rate of zero order reaction is equal to rate constant. i.e. Rate = 2.5 × 10^-4mol L^-1sec^-1.

According to rate law,

2.5 × 10^-4mol L^-1sec^-1 =

i.e. the rate of production of N₂ is 2.5 × 10^-4mol L^-1 sec^-1.

According to rate law,

i.e. rate of formation of H₂ is 3 times rate of reaction = 3 × 2.5 × 10^-4mol L^-1sec^-1

= 7.5 × 10^-4mol L^-1sec^-1

Rate of formation of N₂ and H₂ is 2.5 × 10^-4 mol L^-1sec^-1 and 7.5 × 10^-4 mol L^-1sec^-1 respectively.

Rate of given chemical reaction will be represented as

.

Hence units of rate is bar min^-1

To find units of K, K = rate/[pCH₃OCH₃]^3/2

The unit of k = bar ^-1/2min^-1.

Units of Rate- bar min^-1. and Units of Rate constant K : bar ^-1/2min ^-1

Factors affecting rate of reaction-

1) Temperature (increasing temperature increases rate of reaction)

2) Concentration or pressure of reactants. (Increasing concentration or pressure increases rate of reaction)

3) Presence or absence of a catalyst. (Adding catalyst mostly increases the rate of reaction)

4) The surface area of solid reactant. (If reaction is processing over solid reactant then increasing its surface area increases rate of reaction)

5) Nature of reactants

Let suppose reaction

A ⇒ B;

having rate law, Rate = K [A]²

(i) If the concentration of A is doubled then the rate will affect by the square of concentration i.e. rate will become 2² = 4 times.

(ii) If the concentration of A is halved then the rate will affect the square of concentration i.e (1/2)² = 1/4 times.

(i) Rate becomes 4 times of initial Rate (ii) Rate becomes 1/4 times of initial Rate

Increase in temperature increases the rate constant of a reaction. as we know increase in temperature increases the rate of reaction to satisfy the equation Rate = k [concentration]ⁿ where n can be any real number. k have to increase as concentration is almost not changing over small temperature change. Increasing temperature by 10⁰C almost double the rate constant.

This can be represented quantitively by the help of arrehinus equation-

K = Ae^-Ea/RT, where k is rate constant, E_a is activation energy, R is universal gas constant, T is absolute temperature.

(i) Average rate of reaction over interval is [change in concentration]/[time taken] i.e.

(ii) the pseudo first-order rate constant can be calculated by

K = (2.303/t) log(C_i/C_t) where K is Rate constant,

t is time taken,

C_i is initial concentration

C_t is Concentration at time t.

K = (2.303/30) log (0.55/0.31)

⇒ K = 1.9 × 10^-2 sec^-1

(i) Average rate between 30 to 60 sec is 0.00467 mol L^-1sec^-1

(ii) Pseudo first order rate constant is 1. × 10^-2sec^-1

(i) Order is power raised to reactant in rate law, hence,

Rate = k[A][B]²

(ii) When the concentration of B is increased three times then the rate is affected by the square of reactant. The rate is increased by 9 Times.

(iii) When concentration of reactant both A and B is doubled then the rate will have affected as square of reactant B and Two times of Reactant A. Overall increase in rate is 8 times

(i) When the concentration of B is increased by three times, the rate is increased by nine times.

(ii) When conc. of both reactants doubled then Rate increased 8 times.

When concentration of B is changed then rate of reaction doesn't change that means order with respect to B is 0. But when the concentration of A is doubled rate increased by 2.82 times i.e.2^1.5 = 2.82.Hence order with respect to A is 1.5.

Order with respect to A and B is 1.5 and 0 respectively.

By comparing Experiment I and IV if we increase the concentration of A by 4 times then Rate also increased by 4 times. That means order with respect to A is 1.

By comparing Experiment II and III if we double the concentration of B Rate increases by 4 times that means order with respect to B is 2.

Rate law of reaction will be, Rate = k [A][B]²

To find K, K = rate/[A][B]² i.e. K = 6.0 × 10^-3/[0.1][0.1]²

K = 6 mol^-2L²sec^-1

Order with respect to A and B is 1 and 2 Respectively. And value of K(rate constant) is = 6 mol^-2L²sec^-1

As reaction is first order with respect to A and zero Order with respect to B. Then changing the concentration of B won’t affect the rate of reaction and increasing concentration of A ‘n’ times will increase the rate by ‘n’ times. By this logic lets fill the table- In first blank space concentration of A will be 0.2 mol L^-1 because the rate is doubled. In second blank space, Rate will be 8 × 10^-2mol L^-1min^-1 because the concentration of A is increased 4 Times. In third blank space concentration of A will be 0.1 mol L^-1 because the rate is same as in experiment I.

Half life of first order reaction is, t_1/2 = ln2/K where t_1/2 is half life of first order reaction, K is rate constant of First order reaction.

(i) t_1/2 = ln2/200 s^-1

⇒ t_1/2 = 0.693/200 s^-1 (∵ ln2 = 0.693)

⇒ t_1/2 = 0.003465 sec.

(ii) t_1/2 = ln2/2 min^-1

⇒ t_1/2 = 0.693/2 min^-1 (∵ ln2 = 0.693)

⇒ t_1/2 = 0.3465 min

(iii) t_1/2 = ln2/4 year^-1

⇒ t_1/2 = 0.693/4 years ^-1 (∵ ln2 = 0.693)

⇒ t_1/2 = 0.17325 year.

Half life of 3 reactions are 0.003465 sec, 0.3465 min, 0.17325 year respectively.

Radio active decay occurs via first order rate law,

t_1/2 = 5730 years. rate constant(k) of given decay is 0.693/t_1/2

⇒ 0.693/5730 = 1.2 × 10^-4 year^-1

By first order integrated rate law age of sample will be,

where T is the age of sample A₀ is the initial activity of the sample. and A_t is the activity of sample at any time t.

T = 0.18 × 10⁴ years.

Age of given sample is 0.18 × 10⁴ years.

(i)

(ii) initial conc.is 1.63 × 10^-2 so half conc. is 0.815 × 10^-2M From given information and graph t_1/2 is 1440 sec.

(iii)

(iv) As log [N₂O₅] vs time is straight line given reaction is first order. Hence its rate law will be, Rate = k[N₂O₅]

(v) The slope of above graph is slope = 0.000209

K = 2.303 × slope

⇒ 4.82 × 10^-4sec^-1

Now, t_1/2 = 0.693/K.

⇒ 0.693/4.82 × 10^-4

⇒ t_1/2 = 1438 sec. which is almost equal to (ii)

Given:

Order of the reaction = 1

Let, Initial concentration [R]^°= x

Final concentration [R] = x/16

Rate constant k = 60 s^-1

We know, time

⇒

⇒

Solving, we get t = 4.6 × 10^-2s

Initial concentration, [R]_° = 1μg

Final concentration, [R]

Half-life t_1/2 = 28.1 yrs

Solution:

We know, t_1/2 = 0.693/k

Where, k – rate constant

⇒ k = 0.693/ t_1/2

⇒ k = 0.693/(28.1 yrs)

⇒ k = 0.0246 yrs^-1

Also,

If t = 10yrs, then, using the formula, we get,

⇒

⇒

⇒

⇒

⇒

⇒

⇒ = 0.7824μg

If t = 60yrs, then again, we get,

⇒

⇒

⇒

⇒

⇒ [R]₆₀ = 0.228 μg

∴ 0.2278μg of ⁹⁰Sr will be left after 60 years.

Let, initial concentration be [R]_°

Concentration at 90% completion be ((100-90)/100)×[R]_°

∴ Concentration at 90% be 0.1[R]_°

Concentration at 99% completion be ((100-99)/100)× [R]_°∴ Concentration at 99% be 0.01[R]_°

We know, time

Time taken for 90% completion is

⇒

⇒

Time taken for 99% completion is

⇒

⇒

⇒ t₉₉ = 2t₉₀

Hence, the time taken to complete 9% of the first order reaction is twice the time required for the completion of 90% of the reaction.

Given:

Time t = 40 min

When 30% decomposition is undergone, 70% is the concentration.

We know, time taken

Where, k- rate constant

[R]_° -Initial concentration

[R]-Concentration at time ‘t’

⇒

⇒

⇒

⇒ 258.23 = (2.303/k)

We know, Half-life t_1/2 = 0.693/k

Which can be written as, t_1/2 = 0.3010 × (2.303/k)

⇒ t_1/2 = 0.3010 × 258.23

⇒ t_1/2 = 77.72 min

When t = 0, the total partial pressure is P₀ = 35.0 mm of Hg

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀

Hence, P₀-p = P_t-2( P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time

Where, k- rate constant

[R]_° -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

→ equation 1

At time t = 360 s, P_t = 54 mm of Hg and P₀ = 30 mm of Hg,

Substituting in equation 1,

⇒

⇒ k = 2.175 × 10^-3 s^-1

At time t = 720 s, P_t = 63 mm of Hg and P₀ = 30 mm of Hg,

Substituting in equation 1,

⇒

Thus, k = 2.235 × 10^-3 s^–1

Taking average, k = (2.235 × 10^-3 s^–1 + 2.175 × 10^-3 s^–1)/2

∴ k = 2.21 × 10^-3 s^–1.

When t = 0, the total partial pressure is P₀ = 0.5 atm

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀

Hence, P₀-p = P_t-2(P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time

Where, k- rate constant

[R]_° -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

⇒ → equation 1

At time t = 100 s, P_t = 0.6 atm and P₀ = 0.5 atm,

Substituting in equation 1,

⇒

Thus, k = 2.231 × 10^-3 s^-1

The rate of reaction R = k × P_S02Cl2

When total pressure P_t = 0.65 atm and P₀ = 0.5 atm, then

P_S02Cl2 = 2P₀-P_t

Thus, substituting the values, P_S02Cl2 = 2(0.5)-0.6 = 0.35 atm

R = k × P_S02Cl2 = 2.231 × 10^-3 s^–1 × 0.35

Rate of the reaction R = 7.8 × 10^-4atm s^–1

To convert the temperature in °C to °K we add 273 K.

The graph is given as:

The Arrhenius equation is given by k = Ae^-Ea/RT

Where, k- Rate constant

A- Constant

E_a-Activation Energy

R- Gas constant

T-Temperature

Taking natural log on both sides,

ln k = ln A-(E_a/RT)

→ equation 1

By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –E_a/R.

Slope = (y₂-y₁)/(x₂-x₁)

By substituting the values, slope = -12.301

⇒ –E_a/R = -12.301

But, R = 8.314 JK^-1mol^-1

⇒ E_a = 8.314 JK^-1mol^-1 × 12.301 K

⇒ E_a = 102.27 kJ mol^-1

Substituting the values in equation 1 for data at T = 273K

⇒

⇒

(∵ At T = 273K, ln k = -7.147)

On solving, we get ln A = 37.911

∴ A = 2.91×10⁶

When T = 30⁰C, hence T = 30 + 273 = 303K

⇒

⇒

⇒ ln k = -2.8

⇒ k = 6.08x10^-2s^-1.

When T = 50⁰C, hence T = 50 + 273 = 323K

Substituting in equation 1,

A = 2.91 × 10⁶, E_a = 102.27 kJ mol^-1

Ln K = ln(2.91 × 10⁶)- 102.27/(8.314 × 323)

Thus, ln k = -0.5 and k = 0.607s^-1.

Given,

k = 2.418 × 10^-5 s^-1

T = 546 K

E_a = 179.9 kJ mol^-1 = 179.9 × 10³J mol^-1

The Arrhenius equation is given by k = Ae^-Ea/RT

Taking natural log on both sides,

Ln k = ln A-(E_a/RT)

Substituting the values,

ln(2.418 × 10^-5 ) = ln A-179.9/(8.314 × 546)

ln A = 12.5917

A = 3.9 × 10¹² s^-1(approximately)

Given,

k = 2.0 × 10^–2s^-1

time t = 100s

Concentration [A₀] = 1.0 mol L^-1

We know,

On substituting the values,

Log(1/[A]) = 2.303/2

Log[A] = -2.303/2

[A] = 0.135 mol L^–1

t_1/2 = 3.00 hours

We know, t_1/2 = 0.693/k

∴ k = 0.693/3

k = 0.231hrs^-1

We know, time

Where, k- rate constant

[R]_° -Initial concentration

[R]-Concentration at time ‘t’

Thus, substituting the values,

log([R]₀/[R]) = 0.8

log([R]/[R]₀) = -0.8

[R]/[R]₀ = 0.158

Hence, 0.158 fraction of sucrose remains.

The given equation is

k = (4.5 x 10¹¹ s^-1) e^{-28000 K/T} (i)

Comparing, Arrhenius equation

k = Ae ^-E_a^/RT (ii)

We get, E_a / RT = 28000K / T

⇒ E_a = R x 28000K

= 8.314 J K^-1mol^-1 × 28000 K

= 232792 J mol^–1 or 232.792 kJ mol^–1

We know, The Arrhenius equation is given by k = Ae^-Ea/RT

Taking natural log on both sides,

Ln k = ln A-(E_a/RT)

Thus, log k = log A -(E_a/2.303RT) → eqn 1

The given equation is log k = 14.34 – 1.25 × 10⁴K/T → eqn 2

Comparing 2 equations,

E_a/2.303R = 1.25 × 10⁴K

E_a = 1.25 × 10⁴K × 2.303 × 8.314

E_a = 239339.3 J mol^-1 (approximately)

E_a = 239.34 kJ mol^-1

Also, when t_1/2 = 256 minutes,

k = 0.693 / t_1/2

= 0.693 / 256

= 2.707 × 10^-3 min^-1

k = 4.51 × 10^-5s^–1

Substitute k = 4.51 × 10^-5s^–1 in eqn 2,

log 4.51 × 10^-5 s^–1 = 14.34 – 1.25 × 10⁴K/T

log(0.654-5) = 14.34– 1.25 × 10⁴K/T

T = 1.25 × 10⁴/[ 14.34- log(0.654-5)]

T = 668.9K or T = 669 K

From Arrhenius equation, we obtain

Also, k₁ = 4.5 × 10³ s ^{- 1}

T₁ = 273 + 10 = 283 K

k₂ = 1.5 × 10⁴ s ^{- 1}

E_a = 60 kJ mol ^{- 1} = 6.0 × 10⁴ J mol ^{- 1}

Then,

→ 0.5229 = 3133.627 × (T₂-283)/(283 × T₂)

→ 0.0472T₂ = T₂-283

T₂ = 297K or T₂ = 24⁰ C

We know, time t = (2.303/k) × log([R]₀/[R])

Where, k- rate constant

[R]_° -Initial concentration

[R]-Concentration at time ‘t’

At 298K, If 10% is completed, then 90% is remaining.

t = (2.303/k) × log ([R]₀/0.9[R]₀)

t = (2.303/k) × log (1/0.9)

t = 0.1054 / k

At temperature 308K, 25% is completed, 75% is remaining

t’ = (2.303/k’) × log ([R]₀/0.75[R]₀)

t’ = (2.303/k’) × log (1/0.75)

t’ = 2.2877 / k'

But, t = t’

0.1054 / k = 2.2877 / k'

k' / k = 2.7296

From Arrhenius equation, we obtain

log k₂/k₁ = (E_a / 2.303 R) × (T₂ - T₁) / T₁T₂

Substituting the values,

⇒

E_a = 76640.09 J mol^-1 or 76.64 kJ mol^-1

We know, log k = log A –E_a/RT

Log k = log(4 × 10¹⁰)-(76.64kJ mol^-1/(8.314 × 318))

Log k = -1.986

∴ k = 1.034 x 10^-2 s ^-1

Given, k₂ = 4k₁, T₁ = 293K and T₂ = 313K

We know, From Arrhenius equation, we obtain

⇒

⇒

On solving we get,

E_a = 58263.33 J mol^-1 or 58.26 kJ mol^-1