3 Marks Question

Question 13 Marks

The decomposition of A into product has value of k as $4.5\times10^{3}s^{-1}$ at $10^{\circ}C$ and energy of activation $60~kJ~mol^{-1}$. At what temperature would k be $1.5\times10^{10}s^{-1}$?

Answer

$\log \frac{k_2}{k_1}=\frac{ E _a}{2.303 R }\left[\frac{ T _2- T _1}{T_1 T_2}\right]$
$\log \frac{1.5 \times 10^{10}}{4.5 \times 10^3}=\frac{60000}{2.303 \times 8.314}\left[\frac{T_2-283}{283 T_1}\right]$
or $\quad \log 3.333=3133.62\left[\frac{T_2-283}{283 T_2},\right]$
or $\quad \frac{0.5228}{3133.62}=\frac{ T _2-283}{283 T_2}$
or $\quad 0.0472 T_2= T _2-283$ or $283=0.95278 T_2$
or $\quad T _2=\frac{283}{0.95278}$ or $T _2=297 K$
$\therefore \quad t_2=24^{\circ} C$

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Question 23 Marks

Calculate the half-life of a first order reactions from their rate constants given below :
(a) $200~s^{-1}$
(b) $2~min^{-1}$
(c) $4~year^{-1}$

Answer

(a) $t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{200}=3.465 \times 10^{-3} s^{-1}$
(b) $t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{2}=3.465 \times 10^{-1} min^{-1}$
(c) $t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{4}=1.733 \times 10^{-1}$ year $^{-1}$

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Question 33 Marks

For a reaction $2 H _2 O _2 \frac{ I ^{-}}{\text {alkaline medium }} 2 H _2 O + O _2$
the proposed mechanism is as given below:
(1) $H_{2}O_{2}+I^{-}\rightarrow H_{2}O+IO^{-}$ (slow)
(2) $H_{2}O_{2}+IO^{-}\rightarrow H_{2}O+I^{-}+O_{2}$ (fast)
(i) Write rate law for the reaction.
(ii) Write the overall order of reaction.
(iii) Out of steps (1) and (2), which one is rate determining step?

Answer

(i) The rate law for the reaction is
Rate $=\frac{d\left[ H _2 O _2\right]}{d t}=k\left[ H _2 O _2\right]\left[ I ^{-}\right]$
(ii) This reaction is first order with respect to both $H _2 O _2$ and $I ^{-}$.
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.

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Question 43 Marks

A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minute at 320 K. Calculate the activation energy of the reaction. (Given: log 2=0.3010, log 4=0.6021, $R=8.314$ $J~K^{-1}$ mol).

Answer

$T _2=320 K K _2=\frac{0.693}{20}$
$T _1=300 K K _1=\frac{0.693}{40}$
$\log \frac{ K _2}{K_1}=\frac{ E _a}{2.303 R }\left(\frac{ T _2- T _1}{T_1 T_2}\right)$
$\log \frac{\frac{0.693}{20}}{\frac{0.693}{40}}=\frac{ E _a}{2.303 \times 8.314}\left(\frac{320-300}{320 \times 300}\right)$
$\log \left(\frac{40}{20}\right)=\frac{ E _a}{2.303 \times 8.314}\left(\frac{20}{320 \times 300}\right)$
$E _a=\frac{0.3010 \times 2.303 \times 8.314 \times 320 \times 300}{20}$
$E _{ a }=27.76 kJ$

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Question 53 Marks

The decomposition of $N _2 O _5$ in $CCl _4$ solution follows the first order rate law. The concentration of $N _2 O _5$ measured at different time intervals are given ahead:

Time $(t) \sec$	0	80	160	410	600	1130	1720
$\left[ N _2 O _2\right] mol /$ litre	5.5	5.0	4.8	4.0	3.4	2.4	1.6

Calculate its rate constant at 410 sec and 1130 sec. What does these result show?

Answer

$k=\frac{2.303}{t}=\log \frac{a}{a-x}$
(i) $t=410 sec , \quad k=\frac{2 \cdot 303}{410} \log \frac{5 \cdot 5}{4}$
$=0.0056(1.7404-1.6021)$
$=0.00077448 s^{-1}= 7 . 7 \times 1 0 ^{- 4 } s e c ^{- 1 }$
(ii) $t=1130 sec , k=\frac{2 \cdot 303}{t} \log \frac{a}{a-x}=\frac{2 \cdot 303}{1130} \log \frac{5 \cdot 5}{2 \cdot 4}$
$=0.00204(1.7404-1.3802)$
$=0.00204 \times 0.3602=0.0007348$
$\approx 7.5 \times 10^{-4} sec ^{-1}$
The result shows that this reaction is first order.

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Question 63 Marks

The decomposition of $NH_3$ on platinum surface is zero order reaction. If rate constant (k) is $4\times10^{-3}Ms^{-1},$ how long will it take to reduce the initial concentration of $NH_{3}$ from 0.1 M to 0.064 M.

Answer

The equation for the zero order reaction is:
$[A]=[A_{0}]-kt$ ...(1)
Given:
$[A_{0}]=0.1~M$
$[A]=0.064$
$K=4\times10^{3}M/s$
By putting the value in Eq. (1)
$0.064=0.1-(4\times10^{-3}\times t)$
$0.064-0.1=-(4\times10^{-3}\times t)$
$-0.036=-(4\times10^{-3}\times t)$
$t=\frac{-0.036}{-4\times10^{-3}}=9~s$

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Question 73 Marks

Show that half life of first order reaction is independent to initial concentration.

Answer

$t_{1 / 2} \propto\left[a^0\right]$
or$\quad$ $t_{1 / n} \propto\left[a^0\right]$
$t_{1 / 2} \propto$ constant
Let for any reaction (first order)
$t=t_{1 / 2}$
Then, $\quad a-x=a / 2$
$t=\frac{2.303}{k} \log \frac{(a)}{(a-x)}$
$t_{1 / 2}=\frac{2.303}{k} \log _{10} \frac{a}{a / 2}$
$=\frac{2.303}{k} \log _{10} 2$
$\therefore \quad t_{1 / 2}=\frac{2.303}{k} \times 0.3010=\frac{0.693}{k}$

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Question 83 Marks

How do you differentiate between reaction rate and reaction rate constant. What is meant by the rate controlling step in a reaction?

Answer

Difference between Rate of Reaction and Reaction Rate Constant

Rate of Reaction	Reaction rate Constant
1. Rate of reaction is the change in concentration of a reactant or product per unit time.	1. It is a constant of proportionality in the rate law equation and is equal to the rate of reaction when the molar concentration of each of the reactants is unity.
2. The rate of reaction at any instant of time depends upon the molar concentration of the reactants at that time.	2. The rate constant for a particular reaction at a particular temperature does not depend upon the concentrations of the reactants.
3. Its unit is always moles litre^-1 time^-1	3. Its units depend upon the order of reaction.

The rate controlling step in a reaction is the slowest step on which the rate of reaction depends.

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