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For the reaction A → B, deduce the integrated form of rate law. What will be the nature of the curve when concentration is plotted against time for such reaction?


For the reaction:    A → B.
Let a is the initial concentration of A in g moles L–1 and (a – x) is the concentration in g moles L–1 after time t, then according to law of mass action
          Rate of reaction proportional to (a - x)
or space space space space space space space space space space space space space space space dx over dt proportional to space left parenthesis straight a minus straight x right parenthesis
or space space space space space space space space space space space space space space space dx over dt equals straight k left parenthesis straight a minus straight x right parenthesis
or space space space space space space space space space space space space space space space space fraction numerator dx over denominator left parenthesis straight a minus straight x right parenthesis end fraction equals space straight k space dt
    Integrating the above equation, we get
       negative In space left parenthesis straight a minus straight x right parenthesis space equals space kt plus straight c
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket where space In space is space natural space log right square bracket
when straight t equals 0 comma space space straight x equals 0 comma space straight c equals negative In space straight a
or         negative In left parenthesis straight a minus straight x right parenthesis space equals space kt minus In space straight a
or                 kt equals In fraction numerator straight a over denominator left parenthesis straight a minus straight x right parenthesis end fraction
or                straight k equals 1 over straight t In fraction numerator straight a over denominator left parenthesis straight a minus straight x right parenthesis end fraction
   Changing natural log to base 10, we get
              straight k equals fraction numerator 2.303 over denominator straight t end fraction log fraction numerator straight a over denominator left parenthesis straight a minus straight x right parenthesis end fraction
Nature of the curve: Hypothetical variation of conc. of reactant [R] and product [P] during the course of reaction.

For the reaction:    A → B.Let a is the initial concentration of
Fig. Instantaneous and average rate of a reaction.

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What do you mean by zero order reaction? How the value of rate constant is determined? What is the relation between rate constant and half-life period?

When the rate of the reaction is independent of the concentration of the reactants, the reaction is known as zero order reaction. In zero order reaction, the concentration of reactant (R) remains unaltered during the course of reaction.
It means    rate equals fraction numerator negative straight d open square brackets straight R close square brackets over denominator dt end fraction equals straight k open square brackets straight R close square brackets degree space equals space straight k
or                   straight d open parentheses straight R close parentheses space equals space minus straight k space dt
on integration of this equation, we get
                open square brackets straight R close square brackets space equals space minus kt plus constant          ...(i)
Since        open square brackets straight R close square brackets space equals space open square brackets straight R close square brackets degree
where t = 0 (i.e., no product is formed at the beginning of the reaction), the constant must be zero.
Thus, [R]° = – K x 0 = constant
Constant = [R]°
Put the value of constant in eqn. (i), we get
[R] = – kt + [ R]°                                 ...(ii)
When a plot is drawn between concentration of reactant (R) and time t, a straight line is obtained, slope of which gives the value of – k and intercept on y-axis is equal to the value of [R]°.
Alternatively, the value of k can be obtained from eqn. (i) by putting the value of concentration, [R] at any time t and initial concentration, [R]°.
When the rate of the reaction is independent of the concentration of
Hence,   straight k equals fraction numerator open square brackets straight R close square brackets degree minus open square brackets straight R close square brackets straight t over denominator straight t end fraction
Half life period is the time period to reduce the initial concentration of the reactant to half of its initial value.
Thus, straight t subscript 1 divided by 2 end subscript space equals space fraction numerator open square brackets straight R close square brackets minus begin display style 1 half end style open square brackets straight R close square brackets degree over denominator straight k end fraction space equals fraction numerator open square brackets straight R close square brackets degree over denominator 2 straight k end fraction
Thus, the half life period is directly proportional to the initial concentration of the reactant, [R]° and inversely proportional to the rate constant k.
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Answer the following questions on the above curve for a first order reaction A → P.
(a) What is the relation between slope of this line and rate constant.
(b) (i) Calculate the rate constant of the above reaction if the slope is 2 x 10–4 s–1.
     (ii) Derive the relationship between half life of a first order and its rate constant.


(a) Slope = – k

or k = – 2 x 10–4 mol–1 Ls–1.

(ii) Consider the following first order reaction
A → P

at t = 0 a 0

at t = t (a – x) x

Suppose a is the initial concentration of reactant. After time t, x gm mol lit–1 is changed to P. According to law of mass action the rate of reaction at time t is directly proportional to the concentration of A at that instant i.e., (a – x)

Hence -dxdtα(a-x) or -dxdt=x(a-x) where

is the rate and K is the rate constant, after dxdt rearranging the above equation, we get

-dx(a-x)=k dt        ...(i)
On integrating the above equation

-dx(a-x) = K dt

We get
               -In(a-x) = kt + constant                            dxa-x=ln(a-x) ...(ii)

When t = 0, x = 0 hence – In a = constant substituting this value of constant in eqn. (ii) we get

-In(a-x) = kt-In aor In a - In(a-x) = ktor      Inaa-x = kt                                 In A - In B = InABor            k = 1tIn ax-xor            k = 2.303tlogaa-x.

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For a certain chemical reaction variation in the concentration in [R] vs time(s) plot is given below:
For the reaction write/draw
(i) What is the order of the reaction?
(ii) What are the units of rate constant K?

(iii) Give the relationship between K and t1/2 (half-life period).
(iv) What does the slope of the above line indicate?
(v) Draw the plot Vs. time(s).
Or
For a certain chemical reaction:
A+2B2C+D
The experimentally obtained information is tabulated below:

The experimentally obtained information is tabulated below :

Exp.

[A]0

[B]0

Initial rate of reaction

1

2

3

4

0.30

0.60

0.30

0.60

0.30

0.30

0.60

0.60

0.096

0.384

0.192

0.768

For this reaction:
(i) Derive the order of reaction w.r.t. both the reactants A and B.
(ii) Write the rate law.
(iii) Calculate the value of rate constant K.
(iv) Write the expression for the rate of reaction in terms of A and C. 


(i) First order

(ii) s-1 or min-1

(iii)           t1/2 = 0.693K

(iv)           k=2.303tlogR0R

Where    
           R0 = Initial conc. of the reactant at t = 0R = Concentration of the reactant at time it.              OrLet the law be r = kAxBywhere              x = order with respect to A                        y = order with respect to B

Taking data from experiment (i) to experiment

(iii), we get
0.096 = k(0.30)x (0.30)y ...(ii)
0.384 = k(0.60)x (0.30)y ...(iii)
0.192 = k(0.30)x (0.60)y ...(iv)
0.768 = k(0.60)x (0.60)y ...(v)
Divide eq. (v) by eq. (iv)

0.7080.192 = 2x = 4 = 22               x=2

Divide eq. (iv) by eq. (ii)

0.1920.096 = 2y = 2'             y=1

Substituting the value of x and y in equation (i), we get rate law


              r = kA2 y

(i) Order of reaction  = 2 + 1 = 3

(ii) Rate law;  r=kA2y

(iii) The rate constant K = rA2 B

Substituting the value of r,  A and B from experiment no. (i)

                     k=0.096(0.30)2(0.30) = 0.0960.027

                        = 3.55 mol-2L-2S-1.

(iv) -dCdt=kB2 A rate of reaction in terms of formation of C   -dAdt = kB2A rate of reaction in terms of formation of A

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Ammonia and oxygen react at high temperatures according to the equation
4NH3(g) + 5O2(g) 4NO(g) → + 6H2O(g)
In an experiment, the rate of formation of NO was found to be 3.2 x 10–3 mol L–1 S–1. Calculate the rate of disappearance of ammonia and the rate of formation of water.

Ammonia and oxygen react at high temperatures according to the equation
4NH3(g) + 5O2(g) 4NO(g) → + 6H2O(g)

The relevant rate experessions for the given reaction are

rate=14NH3t       =14NOt=16H2Ot

From the relationship between these expressions
rate of disappearance of NH3 = rate of formation of NO

= 3.2 x 10–3 mol L–1 s–1

Rate of formation of water related to the rate of formation of NO as

                 16H2Ot = 14[NO]t
or                H2Ot = 64NOt                                       = 64×3.2×10-3 mol L-1 s-1                                       = 4.8 × 10-3mol L-1s-1.
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