4 Marks Questions

Question 14 Marks

Mention the factors that affect the rate of a chemical reaction.

Answer

The main factors affecting the rate of chemical reaction are as follows :
(a) Nature of reactants (b) Concentration of reactants (pressure in terms of gases) (c) heat (d) Catalyst.
(i) Concentration of reactants : According to the law of mass proportional reaction, the rate of reaction increases by increasing the concentration of reactants. Expressing the rate of reaction in terms of concentration of reactants is called rate law or rate expression or rate equation.
In gaseous reactions, on increasing the pressure, the rate of the reaction increases in the direction where the number of gaseous molecules is less.
(ii) Nature of reactants : In every reaction, there are different reactants in which the energy of bonds present between them is also different. During the reaction, the bonds of the reactant molecules are broken and new bonds are formed in the products, hence the rate of the reaction depends on the nature of the reactants and the rate of each reaction is different.
(iii) Catalyst : These are the substances which increases the rate of reaction without causing any permanent chemical changes in themselves.

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Question 24 Marks

The decomposition of A into product has value of k as $4.5 \times 10^3 s^{-1}$ at $10^{\circ} C$ and energy of activation $60 kJ mol ^{-1}$. At what temperature would k be $1.5 \times 10^4 s^{-1} ?$

Answer

$\log \frac{ k _2}{ k _1}=\frac{ E _{ a }}{2.303 R }\left[\frac{ T _2- T _1}{T_1 T_2}\right] $
$T _1=10^{\circ} C =10^{\circ} C +273=283 K $
$T _2=?, k _1=4.5 \times 10^3 s^{-1}, k _2=1.5 \times 10^4 s^{-1} $
$E _{ a }=60 kJ mol ^{-1}=60,000 J mol ^{-1} $
$R =8.314 K^{-1} J mol ^{-1}$
On putting values,
$\log \frac{1.5 \times 10^4}{4.5 \times 10^3}=\frac{60,000}{2.303 \times 8.314}\left[\frac{T_2-283}{283 T_2}\right] $
$\log 3.333=\frac{60,000}{19.1471}\left[\frac{T_2-283}{283 T_2}\right]$
$0.5228=3133.63\left[\frac{T_2-283}{283 T_2}\right]$
$\frac{0.5228}{3133.63}=\frac{T_2-283}{283 T_2} $
$1.668 \times 10^{-4} \times 283=\frac{T_2-283}{T_2}$
$0.0472 T_2= T _2-283 $
$T_2-0.0472$ $T_2=283 $
$0.9528 T_2=283$
$ T _2=\frac{283}{0.9528}=297 K $
$T _2=297-273=24^{\circ} C $
Hence, the value of k at $24^{\circ} C$ will be 1.5 × 10^-4 s^-1.

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Question 34 Marks

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with $t_{1 / 2}$ = 3.00 hours. What fraction of sample of surcrose after 8 hours?

Answer

$\underset{\text { Sucrose (in excess) }}{ C _{12} H _{12} O _{11}+ H _2 O } \rightarrow \underset{\text { Glucose }}{ C _6 H _{12} O _6}+\underset{\text { Fructose }}{ C _6 H _{12} O _6}$
This is a first order reaction hence half life for this.
$\begin{aligned} t _{1 / 2} & =\frac{0.693}{ k } \\ k & =\frac{0.693}{ t _{1 / 2}}=\frac{0.693}{3.0 hr } \\ k & =0.231 hr ^{-1}\end{aligned}$
Let the initial concentration of sucrose [R]₀ = 1 mol
$t =8 hr$ and $k =0.231 hr ^{-1}$
$\begin{aligned} k & =\frac{2.303}{ t } \log \frac{[ R ]_0}{[ R ]} \\ 0.231 & =\frac{2.303}{8} \log \frac{1}{[ R ]}\end{aligned}$
$\log \frac{1}{[R]}=\frac{0.231 \times 8}{2.303}=\frac{1.848}{2.303}$
$\log \frac{1}{[R]}=0.8024$
$\frac{1}{[ R ]}=$ Antilog 0.8024
$\frac{1}{[R]}=6.345$
$R=\frac{1}{6.345}=0.1576 M$
Therefore, the amount of remaining sucrose = 0.158 M

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Question 44 Marks

The rate constant for the decomposition of hydrocarbons is 2.418 × 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol. What will be the value of pre-exponential factor?

Answer

$\ln k =-\frac{ E _{ a }}{ RT }+\ln A$
$\ln A =\ln k +\frac{ E _{ a }}{ RT }$
$\log A =\log k +\frac{ E _{ a }}{2.303 RT }$
$\begin{aligned} E _{ a } & =179.9 kJ / mol \\ & =179900 J mol ^{-1} \\ k & =2.418 \times 10^{-5} s^{-1} \\ R & =8.314 Jk ^{-1} \text { and } T =546 K\end{aligned}$
On putting values,
$\log A =\log 2.418 \times 10^{-5}+\frac{179900}{2.303 \times 8.314 \times 546}$
$\log A=\log 10^{-5}+\log 2.418+\frac{179900}{10,454.339}$
$\begin{aligned} \log A & =-5 \log 10+0.3834)+17.208 \\ \log A & =-5+0.3834+17.21 \\ \log A & =-4.6166+17.21 \\ \log A & =12.5934 \\ A & =\text { Antilog } 12.5934\end{aligned}$
$A =3.921 \times 10^{12}$
Hence, prior exponential factor, A = 3.9 × 10¹² s^-1

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