1 Marks Question

Question 11 Mark

The rate constant for a first order reaction is $60~s^{-1}$. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Answer

$t=\frac{2.303}{k} \log \frac{a}{(a-x)}$
if $a=1$ then $(a-x)=1 / 16$
then $\quad t=\frac{2.303}{60} \log \frac{1}{1 / 16}=0.0462 sec$

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Question 21 Mark

The rate constant for first order reaction is 0.0005 $min^{-1}$. Calculate its half-life.

Answer

$t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{0.0005}=1386 min$.

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Question 31 Mark

If in a chemical reaction $A+B \rightarrow Product$ rate low is given by $R=K[A]^{1/2}[B]^{3/2}$ find the order of reaction.

Answer

Order of the Reaction = $\frac{1}{2}+\frac{3}{2}=\frac{4}{2} = 2$

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Question 41 Mark

For the reaction $A\rightarrow B,$ the rate of reaction becomes 27 times when the concentration of A is increased 3 times. What is the order of the reaction?

Answer

$r=k[A]^{n}$ ...(i)
$27r=k[3A]^{n}$ ...(ii)
Dividing (ii) by (i),
$27=3^{n}$ or $3^{3}=3^{n}$ or $n=3.$

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Question 51 Mark

Define threshold energy.

Answer

Threshold Energy: The minimum energy which the colliding molecules must have in order that the collision between them may be effective is called $threshold$$energy$.

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Question 61 Mark

The rate of a reaction becomes four times on increasing temperature from 293 K to 313 K. Find out the activation energy of the reaction assuming it does not change with temperature. ($R=8.314$ J $K^{-1}mol^{-1}$ log 2 = 0.3010)

Answer

$E _a=\frac{2.303 RT _1 T_2}{T_2- T _1} \log \frac{K_2}{K_1}$
$=\frac{2.303 \times 8.314 \times 293 \times 373}{313-293} \times \log 4$
$E _a=52.86 kJ mol ^{-1}$

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Chemistry 1 Marks Question

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