CBSE Class 12 Chemistry Chemical Kinetics Worksheet Set 03

Subjective Questions

Question. For a reaction, \( 2H_2O_2 \xrightarrow{I^- / \text{alkaline medium}} 2H_2O + O_2 \). The proposed mechanism is given below:
(1) \( H_2O_2 + I^- \rightarrow H_2O + IO^- \) (slow)
(2) \( H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2 \) (fast)
(i) Write rate law of the reaction.
(ii) Write overall order of reaction.
(iii) Out of step (1) and (2), which one is rate determining step?
Answer: (i) Rate = \( k [H_2O_2] [I^-] \)
(ii) Order = 2
(iii) Step 1
Detailed Answer:
(i) The rate law for the reaction is \( \text{rate} = -\frac{d[H_2O_2]}{dt} = k[H_2O_2][I^-] \)
(ii) This reaction is first order with respect to both \( H_2O_2 \) and \( I^- \).
1. The overall order of the reaction is bimolecular.
2. The order of the reaction is determined from the slowest step of the reaction mechanism.
(iii) The first reaction is slow, so this is the rate determining step.

Question. For the reaction \( 2N_2O_5 (g) \rightarrow 4NO_2 (g) + O_2 (g) \), the rate of formation of \( NO_2 (g) \) is \( 2.8 \times 10^{-3} \text{ M s}^{-1} \). Calculate the rate of disappearance of \( N_2O_5 (g) \).
Answer: Rate = \( \frac{1}{4} \frac{\Delta[NO_2]}{\Delta t} = -\frac{1}{2} \frac{\Delta[N_2O_5]}{\Delta t} \)
\( \frac{1}{4} (2.8 \times 10^{-3}) = -\frac{1}{2} \frac{\Delta[N_2O_5]}{\Delta t} \)
Rate of disappearance of \( N_2O_5 \) \( \left( -\frac{\Delta[N_2O_5]}{\Delta t} \right) = 1.4 \times 10^{-3} \text{ M/s} \)

Question. For a reaction the rate law expression is represented as follows: Rate = \( k[A][B]^{1/2} \)
(i) Interpret whether the reaction is elementary or complex. Give reason to support your answer.
(ii) Write the unit of rate constant for this reaction if concentration of A and B is expressed in moles/L.
Answer: (i) This is a complex reaction. Order of reaction is a fraction value. i.e., 1.5. Molecularity cannot be a fraction. The reaction occurs in steps, so it is a complex reaction where order and molecularity has different values.
(ii) Rate = \( k[A][B]^{1/2} \)
Unit of Rate constant (k) = \( \frac{\text{Unit of rate}}{\text{Unit of } [A] \times \text{Unit of } [B]^{1/2}} \)
\( = \frac{\text{mol L}^{-1} \text{ s}^{-1}}{(\text{mol L}^{-1})(\text{mol L}^{-1})^{1/2}} \)
\( = \text{mol}^{-1/2} \text{ L}^{1/2} \text{ s}^{-1} \)

Question. The following results have been obtained during the kinetic studies for the reaction: \( P + 2Q \rightarrow R + 2S \)

Determine the rate law expression for the reaction.
Answer: Let the rate law expression be Rate = \( k [P]^x [Q]^y \) from the table we know that
Rate 1 = \( 3.0 \times 10^{-4} = k (0.10)^x (0.10)^y \)
Rate 2 = \( 9.0 \times 10^{-4} = k (0.30)^x (0.30)^y \)
Rate 3 = \( 3.0 \times 10^{-4} = k (0.10)^x (0.30)^y \)
Rate 1/ Rate 3 = \( (1/3)^y \) or \( 1 = (1/3)^y \)
\( \implies y = 0 \)
Rate 2/ Rate 3 = \( (3)^x \) or \( 3 = (3)^x \)
\( \implies x = 1 \)
Rate = \( k [P] \)

Question. For a reaction: \( H_2 + Cl_2 \xrightarrow{hv} 2HCl \), Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.
Answer: (i) Zero order reaction, Molecularity is 2 / bimolecular reaction.
(ii) \( \text{mol L}^{-1} \text{ s}^{-1} \).

Question. For a reaction: \( 2NH_3 (g) \xrightarrow{Pt} N_2(g) + 3H_2(g) \); Rate = k ;
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.
Answer: (i) Zero order, bimolecular / unimolecular.
(ii) \( \text{mol L}^{-1} \text{ s}^{-1} \).

Question. Explain the following terms:
(i) Rate constant (k)
(ii) Half-life period of reaction (\( t_{1/2} \)).
Answer: (i) Rate constant (k): Rate constant is rate of the reaction when the concentration of reactants is unity.
(ii) Half-life period of a reaction (\( t_{1/2} \)): Half-life of a reaction is the time in which the concentration of a reactant is reduced to half of its original value.

Question. Write units of rate constants for zero order and for the second order reactions if the concentration is expressed in \( \text{mol L}^{-1} \) and time in second.
Answer: Zero order: \( \text{mol L}^{-1} \text{s}^{-1} \)
Second order: \( \text{L mol}^{-1} \text{s}^{-1} \)

Question. (i) What is the order of the reaction whose rate constant has same units as the rate of reaction?
(ii) For a reaction \( A + H_2O \rightarrow B \); Rate \( \propto [A] \). What is the order of this reaction?
Answer: (i) Zero order
(ii) Pseudo-first order

Question. (i) Explain why \( H_2 \) and \( O_2 \) do not react at room temperature.
(ii) Write the rate equation for the reaction \( A_2 + 3B_2 \rightarrow 2C \), if the overall order of the reaction is zero.
Answer: (i) Due to high activation energy of the reaction between \( O_2 \) and \( H_2 \).
(ii) Rate = \( k [A_2]^0 [B_2]^0 \)

Question. Derive integrated rate equation for rate constant of a first order reaction.
Answer: \( R \rightarrow P \)
Rate = \( -\frac{d[R]}{dt} = k[R] \)
or \( \frac{d[R]}{[R]} = -k dt \)
Integrating this equation, we get
\( \ln [R] = -kt + I \) …(i)
When \( t = 0 \), \( R = [R]_0 \), where \( [R]_0 \) is the initial concentration of the reactant.
Therefore, equation (i) can be written as
\( \ln [R]_0 = -k \times 0 + I \)
\( \implies \ln [R]_0 = I \)
Substituting the value of I in equation (i)
\( \ln [R] = -kt + \ln[R]_0 \)
Rearranging this equation
\( \ln \frac{[R]}{[R]_0} = -kt \)
\( k = \frac{1}{t} \ln \frac{[R]_0}{[R]} \)
\( k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \)

Question. State a condition under which a bimolecular reaction is kinetically first order reaction.
Answer: Let us take a bimolecular reaction :
\( A + B \rightarrow \text{Product} \)
Rate = \( k [A] [B] \)
When concentration of [B] is taken in excess then rate law will become:
Rate = \( k [A] \)
where, \( k = \text{constant} \)
The rate depends only on one of the reactant as there is negligible change in its concentration so it is bimolecular but is of first order.

Question. The C-14 content of an ancient piece of wood was found to have three tenths of that in living trees. How old is that piece of wood? (log 3 = 0.4771, log 7 = 0.8540, Half-life of C-14 = 5730 years)
Answer: \( k = 0.693 / t_{1/2} \)
\( k = 0.693 / 5730 \text{ years}^{-1} \)
\( t = \frac{2.303}{k} \log \frac{C_0}{C_t} \)
Let \( C_0 = 1 \) and \( C_t = 3/10 \)
so \( C_0/C_t = 1 / (3/10) = 10/3 \)
\( t = \frac{2.303}{0.693} \times 5730 \times \log \frac{10}{3} \)
\( t = 9957 \text{ years} \)

Question. (i) What is the order of the reaction whose rate constant has same units as the rate of reaction?
(ii) For a reaction \( A + H_2O \rightarrow B \); Rate \( \propto [A] \). What is the order of this reaction?
Answer: (i) Zero order
(ii) Pseudo-first order

Question. A first-order reaction has a rate constant \( 1.15 \times 10^{-3} \text{ s}^{-1} \). How long will 5 g of this reactant take to reduce to 3 g?
Answer: Initial amount = 5 g
Final concentration = 3 g
Rate constant = \( 1.15 \times 10^{-3} \text{ s}^{-1} \)
We know that for a First order reaction,
\( t = \frac{2.303}{k} \log \frac{[R_{\text{initial}}]}{[R_{\text{final}}]} \)
\( t = \frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3} \)
\( t = \frac{2.303}{1.15 \times 10^{-3}} \times 0.2219 \)
\( t = 444.379 \text{ s} \)

Question. (i) Why the molecularity of a reaction can not be zero?
(ii) Write the formula for expressing rate of the reaction. \( N_2 + 3H_2 \rightarrow 2NH_3 \)
Answer: (i) The molecularity of a reaction is the number of total molecules taking part in elementary step of reaction. So, minimum one molecule is required for a reaction to occur. Hence, the value of molecularity is never zero.
(ii) \( N_2 + 3H_2 \rightarrow 2NH_3 \)
Rate of reaction = \( -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \)

Question. (i) The rate of reaction decreases with the progress of reaction. Why?
(ii) Reactions having molecularity more than three occur rarely. Why?
Answer: (i) The rate of reaction depends on the concentration of reactants. Since, the concentration of reactants decreases with time, so rate of reaction also decreases.
(ii) It is because of the fact that collision of more than three molecules is not possible at a time.

Question. (i) For which reaction, the rate of reaction does not decrease with time?
(ii) What is the order of photochemical reaction?
Answer: (i) For zero order reaction the rate of reaction does not decrease with time because it does not depend on concentration of reactants.
(ii) Zero Order reaction.

Short Answer Type Questions

Question. For the reaction \( R \rightarrow P \), the concentration of a reactant changes from 0.03 to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer: \( R \rightarrow P \)
\( [R_1] = 0.03 \text{ M} \)
\( [R_2] = 0.02 \text{ M} \)
\( \Delta t = 25 \text{ min} \)
Average rate \( = -\frac{\Delta[R]}{\Delta t} = -\frac{[R_2] - [R_1]}{\Delta t} \)
\( = -\frac{0.02 - 0.03}{25} \)
\( = -\frac{-0.01}{25} \)
\( = 4 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1} \)
Average rate in seconds \( = \frac{4 \times 10^{-4}}{60} \text{ mol L}^{-1} \text{ s}^{-1} \)
\( = 6.66 \times 10^{-6} \text{ mol L}^{-1} \text{ s}^{-1} \)

Question. For the reaction \( 2A + B \rightarrow A_2B \), Rate = \( k[A][B]^2 \), \( k = 2.0 \times 10^{-6} \text{ mol}^{-2} \text{L}^2 \text{s}^{-1} \). Calculate the initial rate of the reaction when [A] = 0.1 \( \text{mol L}^{-1} \) and [B] = 0.2 \( \text{mol L}^{-1} \). Calculate the rate of reaction after [A] is reduced to 0.06 \( \text{mol L}^{-1} \).
Answer: \( 2A + B \rightarrow A_2B \)
Initial rate = \( k[A] [B]^2 \)
\( = 2.0 \times 10^{-6} \times 0.1 \times (0.2)^2 \)
\( = 8.0 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1} \)
When concentration of [A] is reduced from 0.10 \( \text{mol L}^{-1} \) to 0.06 \( \text{mol L}^{-1} \) i.e., 0.04 \( \text{mol L}^{-1} \), concentration of A has been used in the reaction. Therefore, the concentration of B used:
\( = \frac{1}{2} \times 0.04 = 0.02 \text{ mol L}^{-1} \)
Hence, \( [B] = 0.2 - 0.02 = 0.18 \text{ mol L}^{-1} \)
Rate = \( k[A][B]^2 \)
\( = 2.0 \times 10^{-6} \times 0.06 \times (0.18)^2 \)
\( = 3.89 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1} \)

Question. The decomposition of \( NH_3 \) on platinum surface is a zero order reaction. What will be rate of production of \( N_2 \) and \( H_2 \) when the value of k is \( 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \)?
Answer: \( 2NH_3 \rightarrow N_2 + 3H_2 \)
Rate of reaction \( = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt} \)
\( \because \) Reaction is zero order, so, the rate of reaction \( = k[NH_3]^0 = k \)
Rate \( = 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \)
Rate of production of \( N_2 \):
\( \because \) Rate of reaction = Rate of production of \( N_2 \)
\( \implies \) Rate of production of \( N_2 = 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \)
Rate of production of \( H_2 \):
\( \because \) Rate of reaction \( = \frac{1}{3} \times \) Rate of production of \( H_2 \)
\( \implies \) Rate of production of \( H_2 = 3 \times \) Rate of reaction
\( = 3 \times 2.5 \times 10^{-4} \)
\( = 7.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \)

Question. The following data were obtained for the reaction: \( A + 2B \rightarrow C \)

(a) Find the order of reaction with respect to A and B.
(b) Write the rate law and overall order of reaction.
(c) Calculate the rate constant (k).
Answer: (a) Order with respect to A = 2, B = 1.
(b) Rate = \( k[A]^2[B]^1 \); overall order = 3.
(c) Experiment 1: \( 4.2 \times 10^{-2} = k(0.2)^2(0.3) \); \( k = 3.5 \)
Detailed Answer:
Let the order with respect to A be x and B be y.
Rate of reaction = \( k[A]^x[B]^y \)
(i) \( 4.2 \times 10^{-2} = k[0.2]^x[0.3]^y \)
(ii) \( 6.0 \times 10^{-3} = k[0.1]^x[0.1]^y \)
(iii) \( 1.68 \times 10^{-1} = k[0.4]^x[0.3]^y \)
(iv) \( 2.40 \times 10^{-2} = k[0.1]^x[0.4]^y \)
Dividing (iv) by (ii):
\( \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = \frac{k[0.1]^x[0.4]^y}{k[0.1]^x[0.1]^y} \)
\( 4 = (4)^y \implies y = 1 \)
Dividing (i) by (iii):
\( \frac{4.2 \times 10^{-2}}{1.68 \times 10^{-1}} = \frac{k[0.2]^x[0.3]^y}{k[0.4]^x[0.3]^y} \)
\( 0.25 = (0.5)^x \)
\( (0.5)^2 = (0.5)^x \implies x = 2 \)
Rate law = \( k[A]^2[B] \)
Rate constant \( K = \frac{\text{Rate}}{[A]^2[B]} = \frac{6.0 \times 10^{-3}}{(0.1)^2(0.1)} = 6.0 \text{ mol}^{-2} \text{ L}^2 \text{ min}^{-1} \)

Question. Following data are obtained for the reaction: \( N_2O_5 \rightarrow 2NO_2 + 1/2O_2 \)

(a) Show that it follows first order reaction.
(b) Calculate the half-life. (Given log 2 = 0.3010, log 4 = 0.6021)
Answer: (a) \( k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \)
At \( t = 300 \text{ s} \), \( k = \frac{2.303}{300} \log \frac{1.6 \times 10^{-2}}{0.8 \times 10^{-2}} = \frac{2.303}{300} \log 2 = 2.31 \times 10^{-3} \text{ s}^{-1} \)
At \( t = 600 \text{ s} \), \( k = \frac{2.303}{600} \log \frac{1.6 \times 10^{-2}}{0.4 \times 10^{-2}} = \frac{2.303}{600} \log 4 = 2.31 \times 10^{-3} \text{ s}^{-1} \)
k is constant, therefore it follows first order kinetics.
(b) \( t_{1/2} = 0.693/k = 0.693 / (2.31 \times 10^{-3}) = 300 \text{ s} \)

Question. A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)
Answer: \( k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \)
\( 20 \text{ min} = \frac{2.303}{k} \log \frac{100}{75} \) …(i)
\( t = \frac{2.303}{k} \log \frac{100}{25} \) …(ii)
Divide (ii) equation by (i):
\( \frac{t}{20} = \frac{\log 4}{\log 4/3} \)
\( t/20 = 0.6021 / 0.1250 \)
\( \implies t = 96.3 \text{ min} \)

Question. For the first order thermal decomposition reaction, the following data were obtained: \( C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g) \)

Calculate the rate constant (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)
Answer: Given: Initial pressure, \( P_0 = 0.30 \text{ atm} \), \( P_t = 0.50 \text{ atm} \), \( t = 300 \text{ s} \)
Rate constant, \( k = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t} \)
\( = \frac{2.303}{300} \log \frac{0.30}{2(0.30) - 0.50} \)
\( = \frac{2.303}{300} \log \frac{0.30}{0.10} = \frac{2.303}{300} \log 3 \)
\( = \frac{2.303 \times 0.4771}{300} = 0.0036 \text{ s}^{-1} = 3.6 \times 10^{-3} \text{ s}^{-1} \)

Long Answer Type Questions

Question. A reaction is of first order with respect to A and second order with respect to B
(i) Write differential rate equation.
(ii) How is the rate effected when the concentration of B is tripled?
(iii) How is the rate affected when the concentration of both A and B is doubled?
Answer: (i) Differential rate equation: \( \frac{dx}{dt} = k[A][B]^2 \)
(ii) Let \( [A] = a \), \( [B] = b \). \( \text{Rate}_1 = k \times a \times b^2 \).
If [B] is tripled, \( [B] = 3b \). \( \text{Rate}_2 = k \times a \times (3b)^2 = 9 k a b^2 \).
\( \implies \text{Rate}_2 = 9 \times \text{Rate}_1 \). The rate becomes nine times.
(iii) If [A] and [B] are doubled, \( [A] = 2a \), \( [B] = 2b \).
\( \text{Rate}_2 = k(2a)(2b)^2 = 8 k a b^2 \).
\( \implies \text{Rate}_2 = 8 \times \text{Rate}_1 \). The rate becomes eight times.

Question. Reaction \( 2A + B + C \rightarrow D + 2E \) shows first order with respect to A, second order with respect to B and zero order with respect to C. Determine:
(i) Rate law of reaction
(ii) What will be the rate of reaction on doubling of concentration of A, B and C?
(b) For the reaction, \( N_2O_5 \rightarrow 2NO_2 + 1/2O_2 \). The rate of dissociation of \( N_2O_5 \) is \( 5.65 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1} \). Determine:
(i) Rate of reaction
(ii) Rate of formation of \( NO_2 \)
(iii) Rate of formation of \( O_2 \)
Answer: (a)(i) Rate Law: \( \text{Rate } (r) = k[A]^1 [B]^2 [C]^0 \)
(ii) \( r_1 = k[A]^1 [B]^2 [C]^0 \). If concentrations are doubled, \( r_2 = k[2A]^1 [2B]^2 [2C]^0 = 8 k [A]^1 [B]^2 [C]^0 \).
\( \implies r_2 = 8 \times r_1 \). Hence rate becomes eight times.
(b)(i) Rate of reaction = Rate of dissociation of \( N_2O_5 = 5.65 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1} \).
(ii) \( \text{Rate of reaction} = \frac{1}{2} \times \text{Rate of formation of } NO_2 \).
\( \implies \text{Rate of formation of } NO_2 = 2 \times 5.65 \times 10^{-5} = 11.3 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1} \).
(iii) \( \text{Rate of reaction} = 2 \times \text{Rate of formation of } O_2 \).
\( \implies \text{Rate of formation of } O_2 = \frac{1}{2} \times 5.65 \times 10^{-5} = 2.825 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1} \).

Question. (a) A first order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% completed?
(b) Define order of reaction. Write the condition under which a bimolecular reaction follows first order kinetics.
Answer: (a) \( k = \frac{2.303}{40} \log \frac{100}{75} = \frac{2.303}{40} (0.125) = 7.196 \times 10^{-3} \text{ min}^{-1} \).
For 80% completion: \( t = \frac{2.303}{k} \log \frac{100}{20} = \frac{2.303}{7.196 \times 10^{-3}} \log 5 \).
\( t = \frac{2.303 \times 0.6991}{0.007196} = 223.712 \text{ min} \).
(b) Order of reaction: The sum of the coefficients (powers) of the reacting species involved in the rate equation. Condition for pseudo-first order: when one reactant is in large excess.

Question. (i) Write the rate law for a first order reaction. Justify the statement that half life for a first order reaction is independent of the initial concentration of the reactant.
(ii) For a first order reaction, show that the time required for 99% completion of a first order reaction is twice the time required for the completion of 90%.
Answer: (i) \( k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \). At \( t_{1/2} \), \( [R] = [R]_0 / 2 \).
\( k = \frac{2.303}{t_{1/2}} \log \frac{[R]_0}{[R]_0/2} = \frac{2.303}{t_{1/2}} \log 2 \).
\( \implies t_{1/2} = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k} \). This shows \( t_{1/2} \) is independent of \( [R]_0 \).
(ii) \( t_{99\%} = \frac{2.303}{k} \log \frac{100}{1} = \frac{2.303 \times 2}{k} = \frac{4.606}{k} \).
\( t_{90\%} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \).
\( \implies t_{99\%} / t_{90\%} = 2 \). Hence \( t_{99\%} = 2 \times t_{90\%} \).

Question. The following data were obtained for the reaction: \( 2 NO + O_2 \rightarrow 2 NO_2 \)

(i) Find the order of reaction with respect to NO and \( O_2 \).
(ii) Write the rate law and overall order of reaction.
(iii) Calculate the rate constant (k).
Answer: (i) Rate = \( k[NO]^x[O_2]^y \). Comparing Exp 4 and 2: \( \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = \frac{k[0.4]^x[0.1]^y}{k[0.1]^x[0.1]^y} \implies 4 = 4^x \implies x = 1 \).
Comparing Exp 3 and 1: \( \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \frac{k[0.3]^x[0.4]^y}{k[0.3]^x[0.2]^y} \implies 4 = 2^y \implies y = 2 \).
Order w.r.t. NO = 1, w.r.t. \( O_2 = 2 \).
(ii) Rate Law: \( \text{Rate} = k[NO]^1 [O_2]^2 \); Overall order = 3.
(iii) \( k = \frac{\text{Rate}}{[NO][O_2]^2} = \frac{7.2 \times 10^{-2}}{0.3 \times (0.2)^2} = 6.0 \text{ mol}^{-2} \text{ L}^2 \text{ min}^{-1} \).

Question. (i) For a reaction \( A + B \rightarrow P \), the rate is given by Rate = \( k[A] [B]^2 \)
(a) How is the rate of reaction affected if the concentration of B is doubled ?
(b) What is the overall order of reaction if A is present in large excess ?
(ii) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (log 2 = 0.3010)
Answer: (i)(a) If [B] is doubled, Rate = \( k[A][2B]^2 = 4 k [A][B]^2 \). Rate becomes 4 times.
(b) If A is in excess, rate depends only on B. Rate = \( k'[B]^2 \). Order is 2.
(ii) \( t_{1/2} = 30 \text{ min} = 1800 \text{ s} \). \( k = \frac{2.303 \times 0.3010}{1800} \).
\( t_{90\%} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \).
Substituting k: \( t_{90\%} = \frac{2.303 \times 1800}{2.303 \times 0.3010} = 5980 \text{ s} = 5.98 \times 10^3 \text{ s} \).

Question. For a given chemical reaction determine the order of reaction with respect to \( HgCl_2 \), with respect to \( C_2O_4^{2-} \) and overall reaction. \( HgCl_2(aq) + C_2O_4^{2-}(aq) \rightarrow 2 Cl^-(aq) + 2 CO_2(g) + Hg_2Cl_2(s) \)

Answer: Rate = \( k[HgCl_2]^m[C_2O_4^{2-}]^n \). Comparing Exp 3 and 4: \( \frac{7.1 \times 10^{-5}}{8.9 \times 10^{-6}} = \frac{k[0.052]^m[0.30]^n}{k[0.052]^m[0.15]^n} \implies 8 = 2^n \implies n = 3 \). (Correction based on calculation: \( 7.1 \times 10^{-5} / 1.8 \times 10^{-5} \approx 4 = 2^n \implies n = 2 \)).
Comparing Exp 4 and 1: \( \frac{8.9 \times 10^{-6}}{1.8 \times 10^{-5}} = \frac{k[0.052]^m[0.15]^n}{k[0.105]^m[0.15]^n} \implies 0.493 = (0.495)^m \implies m = 1 \).
Order w.r.t. \( HgCl_2 = 1 \), w.r.t. oxalate = 2. Overall order = 3.