Chemical Kinetics and Nuclear Chemistry
Explanation:
The given problem involves calculating the age of a wood sample using the carbon-14 dating method. Carbon-14 ($^{14}C$) is a radioactive isotope of carbon that decays over time, and its ratio compared to carbon-12 ($^{12}C$) can be used to date organic materials. The ratio of $\frac{^{14}C}{^{12}C}$ in the wood is $\frac{1}{8}$th of that in the atmosphere, suggesting the $^{14}C$ has decayed. The half-life of $^{14}C$ is given as 5730 years, which is the time for half the $^{14}C$ to decay.
To solve for the age of the wood, we use the formula for radioactive decay:
$N = N_0 e^{-\lambda t},$where $N$ is the remaining amount of $^{14}C$, $N_0$ is the original amount of $^{14}C$, $\lambda$ is the decay constant, and $t$ is the time (age of the wood).
Since the ratio $\frac{N}{N_0} = \frac{1}{8}$, we infer that $N = \frac{N_0}{8}$. Substituting into the decay equation and solving for $t$, first we find the decay constant $\lambda$:
$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730}.$Substitute $\lambda$ and rearrange to solve for $t$:
$t = \frac{\ln(8)}{\lambda} = \frac{\ln(8)}{0.693} \times 5730 = 3 \times 5730 = 17190 \text{ years}.$This calculation shows that the wood sample is 17190 years old.
$\mathrm{r}=\mathrm{k}[\mathrm{A}]$ for a reaction, $50 \%$ of $\mathrm{A}$ is decomposed in 120 minutes. The time taken for $90 \%$ decomposition of $\mathrm{A}$ is _________ minutes.
Explanation:
$\mathrm{r}=\mathrm{k}[\mathrm{A}]$
So, order of reaction $=1$
$\mathrm{t}_{1 / 2}=120 \mathrm{~min}$
For $90 \%$ completion of reaction
$\begin{aligned} & \Rightarrow \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right) \\ & \Rightarrow \frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{2.303}{\mathrm{t}} \log \frac{100}{10} \\ & \therefore \mathrm{t}=399 \mathrm{~min} . \end{aligned}$
$\mathrm{NO}_2$ required for a reaction is produced by decomposition of $\mathrm{N}_2 \mathrm{O}_5$ in $\mathrm{CCl}_4$ as by equation
$2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
The initial concentration of $\mathrm{N}_2 \mathrm{O}_5$ is $3 \mathrm{~mol} \mathrm{~L}^{-1}$ and it is $2.75 \mathrm{~mol} \mathrm{~L}^{-1}$ after 30 minutes.
The rate of formation of $\mathrm{NO}_2$ is $\mathrm{x} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$, value of $\mathrm{x}$ is _________. (nearest integer)
Explanation:
Rate of reaction (ROR)
$\begin{aligned} & =-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=\frac{1}{4} \frac{\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta \mathrm{t}} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=-\frac{1}{2} \frac{(2.75-3)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{(-0.25)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \text { ROR }=\frac{1}{240} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \end{aligned}$
Rate of formation of $\mathrm{NO}_2=\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=4 \times \mathrm{ROR}$
$=\frac{4}{240}=16.66 \times 10^{-3} \mathrm{molL}^{-1} \mathrm{~min}^{-1} \simeq 17 \times 10^{-3} \text {. }$
The rate of First order reaction is $0.04 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ at 10 minutes and $0.03 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ at 20 minutes after initiation. Half life of the reaction is _______ minutes.
(Given $\log 2=0.3010, \log 3=0.4771$)
Explanation:
$\begin{aligned} & 0.04=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 10 \times 60} \quad \text{..... (1)}\\ & 0.03=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 20 \times 60} \quad \text{..... (2)} \end{aligned}$
$\begin{aligned} \frac{4}{3} & =\mathrm{e}^{600 \mathrm{k}(2-1)} /(2) \\ \frac{4}{3} & =\mathrm{e}^{600 \mathrm{k}} \\ \ln \frac{4}{3} & =600 \mathrm{k} \\ \ln \frac{4}{3} & =600 \times \frac{\ln 2}{\mathrm{t}_{1 / 2}} \\ \mathrm{t}_{1 / 2} & =600 \frac{\ln 2}{\ln \frac{4}{3}} \sec \\ \mathrm{t}_{1 / 2} & =600 \times \frac{\log 2}{\log 4-\log 3} \mathrm{sec} .=10 \times \frac{0.3010}{0.6020-0.477} \mathrm{~min} \\ \mathrm{t}_{1 / 2} & =24.08 \mathrm{~min} \end{aligned}$
Ans. 24
The half-life of radioisotope bromine - 82 is 36 hours. The fraction which remains after one day is ________ $\times 10^{-2}$.
(Given antilog $0.2006=1.587$)
Explanation:
Half life of bromine $-82=36$ hours
$\begin{aligned} & \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{~K}=\frac{0.693}{36}=0.01925 \mathrm{~hr}^{-1} \\ & 1^{\text {st }} \text { order rxn kinetic equation } \\ & \mathrm{t}=\frac{2.303}{\mathrm{~K}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\ & \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\frac{\mathrm{t} \times \mathrm{K}}{2.303}(\mathrm{t}=1 \text { day }=24 \mathrm{~hr}) \\ & \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\frac{24 \mathrm{hr} \times 0.01925 \mathrm{hr}^{-1}}{2.303} \\ & \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=0.2006 \\ & \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\mathrm{anti} \log (0.2006) \\ & \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=1.587 \\ & \text { If }=1 \\ & \frac{1}{1-\mathrm{x}}=1.587 \Rightarrow 1-\mathrm{x}=0.6301=\text { Fraction remain } \\ & \mathrm{after} \text { one day } \end{aligned}$
For a reaction taking place in three steps at same temperature, overall rate constant $\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3}$. If $\mathrm{Ea}_1, \mathrm{Ea}_2$ and $\mathrm{Ea}_3$ are 40, 50 and $60 \mathrm{~kJ} / \mathrm{mol}$ respectively, the overall $\mathrm{Ea}$ is ________ $\mathrm{kJ} / \mathrm{mol}$.
Explanation:
$\begin{aligned} & \mathrm{K}=\frac{\mathrm{K}_1 \cdot \mathrm{K}_2}{\mathrm{~K}_3}=\frac{\mathrm{A}_1 \cdot \mathrm{A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}} \\ & \mathrm{~A} \cdot \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}=\frac{\mathrm{A}_1 \mathrm{~A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}}\end{aligned}$
$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}=40+50-60=30 \mathrm{~kJ} / \mathrm{mole} .$
Time required for completion of $99.9 \%$ of a First order reaction is ________ times of half life $\left(t_{1 / 2}\right)$ of the reaction.
Explanation:
$\frac{\mathrm{t}_{99.9 \%}}{\mathrm{t}_{1 / 2}}=\frac{\frac{2.303}{\mathrm{k}}\left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)}{\frac{2.303}{\mathrm{k}} \log 2}=\frac{\log \left(\frac{100}{100-99.9}\right)}{\log 2}=\frac{\log 10^3}{\log 2}=\frac{3}{0.3}=10$
Consider the following data for the given reaction
$2 \mathrm{HI}_{(\mathrm{g})} \rightarrow \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})}$
The order of the reaction is _________.
Explanation:
Let, $\mathrm{R}=\mathrm{k}[\mathrm{HI}]^{\mathrm{n}}$
using any two of given data,
$\begin{aligned} & \frac{3 \times 10^{-3}}{7.5 \times 10^{-4}}=\left(\frac{0.01}{0.005}\right)^{\mathrm{n}} \\\\ & \mathrm{n}=2 \end{aligned}$
For a reaction $A \xrightarrow{\mathrm{K}_1} \mathrm{~B} \xrightarrow{\mathrm{K}_2} \mathrm{C}$ If the rate of formation of B is set to be zero then the concentration of B is given by :
Integrated rate law equation for a first order gas phase reaction is given by (where $\mathrm{P}_{\mathrm{i}}$ is initial pressure and $\mathrm{P}_{\mathrm{t}}$ is total pressure at time $t$)
The activation energy of the catalysed backward reaction is ___________ $\mathrm{kJ}~ \mathrm{mol}^{-1}$.
Explanation:
$k = Ae^{-E_a/RT}$
where $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature in Kelvin.
The equation is used for both the uncatalyzed forward reaction and the catalyzed backward reaction. By setting the two equations equal to each other and cancelling out the pre-exponential factor, we get:
$e^{\frac{300 \times 10^3}{600 \times R}} = e^{\frac{-E_a}{300 \times R}}$
Simplifying the equation, we get:
$\frac{300 \times 10^3}{600 \times R} = \frac{E_a}{300 \times R}$
Solving for $E_a$, we get:
$E_a = \frac{10^3}{2}\times 300 = 150 \times 10^3\ \mathrm{J~mol^{-1}} = 150\ \mathrm{kJ~mol^{-1}}$
This gives us the activation energy for the uncatalyzed forward reaction.
To find the activation energy for the catalyzed backward reaction, we use the relationship:
$E_{\text{rev,catalysed}} = E_{\text{fwd,uncat}} - \Delta H_{\text{forward reaction}}$
Substituting the given values, we get:
$E_{\text{rev,catalysed}} = 150\ \mathrm{kJ~mol^{-1}} - 20\ \mathrm{kJ~mol^{-1}} = 130\ \mathrm{kJ~mol^{-1}}$
Therefore, the activation energy for the catalyzed backward reaction is $\boxed{130\ \mathrm{kJ~mol^{-1}}}$.
A(g) $\to$ 2B(g) + C(g) is a first order reaction. The initial pressure of the system was found to be 800 mm Hg which increased to 1600 mm Hg after 10 min. The total pressure of the system after 30 min will be _________ mm Hg. (Nearest integer)
Explanation:
$800 + 2x = 1600$
$2x = 800$
$x = 400$
The rate constant (k) can be found as:
$k = \frac{2.303}{10} \log \frac{800}{400} = \frac{2.303 \times \log 2}{10}$
For 30 minutes, we can set up the equation:
$k = \frac{2.303}{30} \log \frac{800}{800 - y}$
$\frac{2.303 \times \log 2}{10} = \frac{2.303}{30} \log \left(\frac{800}{800 - y}\right)$
Solving for y:
$\left(\frac{800}{800 - y}\right) = 8$
$800 - y = 100$
$y = 700$
Now we can find the total pressure after 30 minutes:
Total pressure = (800 - y) + 2y + y
Total pressure = 100 + 1400 + 700
Total pressure = 2200 mm Hg
So, the total pressure of the system after 30 minutes is 2200 mm Hg.
$\mathrm{t}_{87.5}$ is the time required for the reaction to undergo $87.5 \%$ completion and $\mathrm{t}_{50}$ is the time required for the reaction to undergo $50 \%$ completion. The relation between $\mathrm{t}_{87.5}$ and $\mathrm{t}_{50}$ for a first order reaction is $\mathrm{t}_{87.5}=x \times \mathrm{t}_{50}$ The value of $x$ is ___________. (Nearest integer)
Explanation:
$\ln\left(\frac{1}{1-p}\right) = kt$
For tâ‚…â‚€ (50% completion), p = 0.5:
$\ln\left(\frac{1}{1-0.5}\right) = k\cdot t_{50}$
$\ln\left(\frac{1}{0.5}\right) = k\cdot t_{50}$
$\ln(2) = k\cdot t_{50}$
For t₈₇.₅ (87.5% completion), p = 0.875:
$\ln\left(\frac{1}{1-0.875}\right) = k\cdot t_{87.5}$
$\ln\left(\frac{1}{0.125}\right) = k\cdot t_{87.5}$ $\ln(8) = k\cdot t_{87.5}$
Now, we need to find the relationship between t₈₇.₅ and t₅₀:
$\frac{k\cdot t_{87.5}}{k\cdot t_{50}} = \frac{\ln(8)}{\ln(2)}$
Since the k's cancel out, we have:
$\frac{t_{87.5}}{t_{50}} = \frac{\ln(8)}{\ln(2)}$
Using the property of logarithms, we get:
$\frac{t_{87.5}}{t_{50}} = \frac{\ln(2^3)}{\ln(2)}$
$\frac{t_{87.5}}{t_{50}} = 3$
So, the value of x is 3.
The reaction $2 \mathrm{NO}+\mathrm{Br}_{2} \rightarrow 2 \mathrm{NOBr}$
takes places through the mechanism given below:
$\mathrm{NO}+\mathrm{Br}_{2} \Leftrightarrow \mathrm{NOBr}_{2}$ (fast)
$\mathrm{NOBr}_{2}+\mathrm{NO} \rightarrow 2 \mathrm{NOBr}$ (slow)
The overall order of the reaction is ___________.
Explanation:
The overall order of a reaction is determined by the slow (rate-determining) step.
Here, the slow step is: $ \text{NOBr}_2 + \text{NO} \rightarrow 2 \text{NOBr} $
This is a second-order reaction: first-order with respect to $\text{NOBr}_2$ and first-order with respect to NO.
However, $\text{NOBr}_2$ is not a reactant in the overall reaction. It's an intermediate. So, we need to express it in terms of the initial reactants.
From the first (fast) step, we get: $ \text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2 $
In the steady state approximation, the rate of formation of $\text{NOBr}_2$ equals the rate of its consumption. We can write this as:
$ k_1[\text{NO}][\text{Br}_2] = k_{-1}[\text{NOBr}_2] + k_2[\text{NOBr}_2][\text{NO}] $
Since the second reaction (slow step) is much slower than the first one, $k_2[\text{NOBr}_2][\text{NO}]$ term is negligible in comparison to $k_{-1}[\text{NOBr}_2]$.
So, we can simplify to:
$ k_1[\text{NO}][\text{Br}_2] \approx k_{-1}[\text{NOBr}_2] $
Solving for $[\text{NOBr}_2]$, we get: $ [\text{NOBr}_2] \approx \frac{k_1}{k_{-1}} [\text{NO}][\text{Br}_2] $
Substitute $[\text{NOBr}_2]$ into the rate equation for the slow step:
$ \text{Rate} = k_2[\text{NO}][\text{NOBr}_2] $
$ \text{Rate} = k_2[\text{NO}] \left( \frac{k_1}{k_{-1}} [\text{NO}][\text{Br}_2] \right) $
So the overall reaction order is 3 (2 with respect to NO and 1 with respect to $\text{Br}_2$).
$\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{H}_{2} \mathrm{O}$
The above reaction was studied at $300 \mathrm{~K}$ by monitoring the concentration of $\mathrm{FeSO}_{4}$ in which initial concentration was $10 \mathrm{M}$ and after half an hour became 8.8 M. The rate of production of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is _________ $\times 10^{-6} \mathrm{~mol} \mathrm{~L} \mathrm{~s}^{-1}$ (Nearest integer)
Explanation:
$\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{H}_{2} \mathrm{O}$
The rates of reaction can be written as :
$ \begin{aligned} & \mathrm{ROR}=-\frac{\Delta\left[\mathrm{KClO}_3\right]}{\Delta \mathrm{t}}=\frac{-1}{6} \frac{\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}} \\\\ &=\frac{+1}{3} \frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}} \\ \end{aligned} $
From this, we can express the rate of formation of $\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$ in terms of the change in concentration of $\mathrm{FeSO}_4$ :
$ \frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{-\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}} $
We know that the initial concentration of $\mathrm{FeSO}_4$ was 10 M and after 30 minutes (or 1800 seconds), it became 8.8 M. Substituting these values in, we get :
$=\frac{1}{2} \frac{(10-8.8)}{30 \times 60} = 0.333 \times 10^{-3} $
To express the rate in terms of $10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$, we multiply the rate by $10^{3}$ :
$ =333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$
Therefore, the rate of production of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is $333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}$.
The number of incorrect statement/s from the following is ___________
A. The successive half lives of zero order reactions decreases with time.
B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction
C. Order and molecularity of a chemical reaction can be a fractional number
D. The rate constant units of zero and second order reaction are $\mathrm{mol} ~\mathrm{L}^{-1} \mathrm{~s}^{-1}$ and $\mathrm{mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}$ respectively
Explanation:
A. The successive half lives of zero order reactions decreases with time. This statement is correct. The half-life of a zero-order reaction is given by the formula $t_{1/2} = [A]_0/2k$, where $[A]_0$ is the initial concentration and k is the rate constant. As the concentration decreases, the half-life also decreases.
B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction. This statement is correct. For example, in a zero-order reaction, the rate of reaction does not depend on the concentration of reactants.
C. Order and molecularity of a chemical reaction can be a fractional number. This statement is incorrect. The order of a reaction can be zero, fractional or negative. However, molecularity, which refers to the number of reactant molecules taking part in an elementary reaction, can only be a whole number.
D. The rate constant units of zero and second order reaction are $\mathrm{mol} ~\mathrm{L}^{-1} \mathrm{~s}^{-1}$ and $\mathrm{mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}$ respectively. This statement is correct. The units of the rate constant for a zero-order reaction are M/s or mol·L−1·s−1 . The units of the rate constant for a second-order reaction are 1/(M·s) or L·mol−1·s−1 .
A molecule undergoes two independent first order reactions whose respective half lives are 12 min and 3 min. If both the reactions are occurring then the time taken for the 50% consumption of the reactant is ___________ min. (Nearest integer)
Explanation:
In the given problem, a molecule is undergoing two separate, simultaneous first-order reactions. Each of these reactions has its own rate constant and half-life. The half-life ($T$) of a first-order reaction is related to its rate constant ($k$) by the equation:
$T = \frac{\ln(2)}{k}$
When two or more first-order reactions are occurring independently and simultaneously (also known as parallel or competing reactions), the overall rate constant for the disappearance of the reactant is the sum of the rate constants for the individual reactions:
$k_{\text{total}} = k_1 + k_2$
Conversely, the overall half-life for the disappearance of the reactant is determined by the reciprocal of the sum of the reciprocals of the individual half-lives:
$\frac{1}{T_{\text{total}}} = \frac{1}{T_1} + \frac{1}{T_2}$
This equation reflects the fact that the reactant is disappearing more quickly than it would due to any single reaction, because it is being consumed by two reactions at once.
Substituting the given half-lives of $12$ minutes and $3$ minutes into this equation gives:
$\begin{align} \frac{1}{T_{\text{total}}} &= \frac{1}{12\text{ min}} + \frac{1}{3\text{ min}} \\\\ &= \frac{5}{12\text{ min}} \end{align}$
Solving for $T_{\text{total}}$ then gives:
$T_{\text{total}} = \frac{12}{5}\text{ min} = 2.4\text{ min}$
This is the time taken for $50\%$ of the reactant to be consumed when both reactions are occurring. Rounding to the nearest integer gives an answer of $2$ minutes.
The number of given statement/s which is/are correct is __________.
(A) The stronger the temperature dependence of the rate constant, the higher is the activation energy.
(B) If a reaction has zero activation energy, its rate is independent of temperature.
(C) The stronger the temperature dependence of the rate constant, the smaller is the activation energy.
(D) If there is no correlation between the temperature and the rate constant then it means that the reaction has negative activation energy.
Explanation:
Clearly, if $\mathrm{E}_a=0, \mathrm{~K}$ is temperature independent if $\mathrm{E}_a>0, \mathrm{~K}$ increase with increase in temperature if $\mathrm{E}_a<0, \mathrm{~K}$ decrease with increase in temperature
- Rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy.
- Higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant.
- The pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.
a. A high activation energy usually implies a slow reaction.
b. $k=\mathrm{P} \times \mathrm{Z} \times \mathrm{e}^{-\mathrm{E}_a / \mathrm{RT}}$
c. The pre-exponential factor $(\mathrm{A}=\mathrm{P} \times \mathrm{Z})$ is independent of the activation energy and the energy of molecules.
$\mathrm{A}$ $\rightarrow \mathrm{B}$
The above reaction is of zero order. Half life of this reaction is $50 \mathrm{~min}$. The time taken for the concentration of $\mathrm{A}$ to reduce to one-fourth of its initial value is ____________ min. (Nearest integer)
Explanation:
$ \mathrm{a}-\mathrm{x}=\frac{\mathrm{a}}{4} \Rightarrow \mathrm{x}=\frac{3 \mathrm{a}}{4} $
$ \mathrm{t}_{1 / 2}=\frac{\mathrm{a}}{2 \mathrm{~K}}=50 \mathrm{~min} .$
$ \Rightarrow \frac{\mathrm{a}}{\mathrm{K}}=100 \mathrm{~min} . $
$ \mathrm{t}=\frac{\mathrm{x}}{\mathrm{K}}=\frac{3 \mathrm{a}}{4 \mathrm{~K}}=75 \mathrm{~min} $
A and B are two substances undergoing radioactive decay in a container. The half life of A is 15 min and that of B is 5 min. If the initial concentration of B is 4 times that of A and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? _____________ min.
Explanation:
After $15 \min ,[A]=[B]=\frac{Co}{2}$
(Given : $\ln 10=2.303$ and $ \log 2=0.3010 \text { )}$
Explanation:
$ \mathrm{t}_{1 / 2}=\frac{0.6932}{20}=\frac{\ln 2}{20} $
Required time $=\mathrm{n} \times \mathrm{t}_{1 / 2}$
$C = {{{C_0}} \over {{2^n}}} = {{{C_0}} \over {32}}$
$ \Rightarrow $ ${2^n} = 32$ = ${2^5}$
$ \Rightarrow $ n = 5
$ \begin{aligned} \text { Required time } & =\frac{5 \times 0.6932}{20} \\\\ & =0.173 \mathrm{~min} \\\\ & =17.3 \times 10^{-2} \mathrm{~min} \end{aligned} $
A $\to$ B
The rate constants of the above reaction at 200 K and 300 K are 0.03 min$^{-1}$ and 0.05 min$^{-1}$ respectively. The activation energy for the reaction is ___________ J (Nearest integer)
(Given : $\mathrm{ln10=2.3}$
$\mathrm{R=8.3~J~K^{-1}~mol^{-1}}$
$\mathrm{\log5=0.70}$
$\mathrm{\log3=0.48}$
$\mathrm{\log2=0.30}$)
Explanation:
Given: $\ln 10=2.3 ; \log 2=0.3$
Explanation:
$ k=\frac{1}{t} \ln \frac{a}{a-x} $
$ \Rightarrow $ $ t=\frac{1}{k} \ln \frac{a}{a-x} $
When reaction is $60 \%$ completed,
$ x=\frac{60}{100} a=0.6 a, t=540 \text { seconds } ; $
$k= \frac{1}{t_1} \ln \frac{a}{a-0.6 a}$
$ \therefore $ $t_1=\frac{1}{k} \ln \frac{a}{0.4 a} $
When reaction is $90 \%$ completed, i.e., $x=0.9 a$
$ \begin{aligned} k= \frac{1}{t_2} \ln \frac{a}{a-0.9 a} \\\\ \Rightarrow t_2=\frac{1}{k} \ln \frac{a}{0.1 a} \\\\ \therefore \frac{t_1}{t_2} =\frac{\frac{1}{k} \ln \frac{a}{0.4 a}}{\frac{1}{k} \ln \frac{a}{0.1 a}} \\\\ \Rightarrow \frac{540}{t_2} =\frac{\ln \frac{10}{4}}{\ln 10} \end{aligned} $
$ \Rightarrow $ $\frac{540}{t_2} =\frac{2.3\log 10-2.3\log 4}{2.3\log 10}$
$ \Rightarrow $ $\frac{540}{t_2} =\frac{2.3-2.3(0.6)}{2.3}$
$ \Rightarrow $ $\frac{540}{t_2} =0.4$
$ \Rightarrow $ $ t_2=\frac{540}{0.4}=1350 $ sec
If compound A reacts with B following first order kinetics with rate constant $2.011 \times 10^{-3} \mathrm{~s}^{-1}$. The time taken by $\mathrm{A}$ (in seconds) to reduce from $7 \mathrm{~g}$ to $2 \mathrm{~g}$ will be ___________. (Nearest Integer)
$[\log 5=0.698, \log 7=0.845, \log 2=0.301]$
Explanation:
$t = {{2.303} \over k}\log {{{C_0}} \over {{C_t}}}$
$ = {{2.303} \over {2.011 \times {{10}^{ - 3}}}}\log {7 \over 2}$
$ = {{2.303 \times {{10}^3}} \over {2.011}}(.845 - .301)$
$ = 622.99$
$ \approx 623$ sec.
For conversion of compound A $\to$ B, the rate constant of the reaction was found to be $\mathrm{4.6\times10^{-5}~L~mol^{-1}~s^{-1}}$. The order of the reaction is ____________.
Explanation:
As unit is L mol$^{-1}$ s$^{-1}$, order of the reaction is 2.
For certain chemical reaction $X\to Y$, the rate of formation of product is plotted against the time as shown in the figure. The number of $\mathrm{\underline {correct} }$ statement/s from the following is ___________.
(A) Over all order of this reaction is one.
(B) Order of this reaction can't be determined.
(C) In region I and III, the reaction is of first and zero order respectively.
(D) In region-II, the reaction is of first order.
(E) In region-II, the order of reaction is in the range of 0.1 to 0.9.
Explanation:
In region I and II, slope of the graph is positive so reaction is nagative order.
In region III, slope of the graph is zero so the reaction is of zero order.
$ \therefore $ Order of this reaction can't be determined. So only (B) is correct.
A first order reaction has the rate constant, $\mathrm{k=4.6\times10^{-3}~s^{-1}}$. The number of correct statement/s from the following is/are __________
Given : $\mathrm{\log3=0.48}$
A. Reaction completes in 1000 s.
B. The reaction has a half-life of 500 s.
C. The time required for 10% completion is 25 times the time required for 90% completion.
D. The degree of dissociation is equal to ($\mathrm{1-e^{-kt}}$)
E. The rate and the rate constant have the same unit.
Explanation:
$\mathrm{k}=4.6 \times 10^{-3} \mathrm{~s}^{-1}$
$\mathrm{kt}=\ln \frac{1}{1-\alpha}$
$\alpha=1-\mathrm{e}^{-\mathrm{kt}}$
Reaction completes at infinite time.
(B) For first order reaction,
Half-life $=\frac{0.693}{4.6 \times 10^{-3}}=150.65 \mathrm{~s}$
(C) For $10 \%$ completion, $t=t_1$
$ \begin{aligned} & a=100, a-x=100-10=90 \\\\ & k=\frac{2.303}{t_1} \log \frac{100}{90}=4.6 \times 10^{-3} \\\\ & t_1=\frac{2.303}{4.6 \times 10^{-3}} \times 0.04575 \end{aligned} $
For $90 \%$ completion $t=t_2, a=100, a-x=100-90=10$
$ \begin{aligned} & k=\frac{2.303}{t_2} \log \frac{100}{10}=4.6 \times 10^{-3} \\\\ & \frac{2.303}{t_2} \times 1=4.6 \times 10^{-3} \Rightarrow t_2=\frac{2.303}{4.6 \times 10^{-3}} \\\\ & \frac{t_2}{t_1}=\frac{1}{0.04575}=21.85 \quad \text { (Incorrect) } \end{aligned} $
(E) Unit of rate = mol L–1s–1
Unit of rate constant = s–1
Both have different units.
$\therefore$ Number of correct statements $=1$
For the first order reaction A $\to$ B, the half life is 30 min. The time taken for 75% completion of the reaction is _________ min. (Nearest integer)
Given : log 2 = 0.3010
log 3 = 0.4771
log 5 = 0.6989
Explanation:
$=2 \times \mathrm{t}_{1 / 2}$
$=2 \times 30$
$=60 \mathrm{~min}$
The number of correct statement/s from the following is __________
A. Larger the activation energy, smaller is the value of the rate constant.
B. The higher is the activation energy, higher is the value of the temperature coefficient.
C. At lower temperatures, increase in temperature causes more change in the value of k than at higher temperature
D. A plot of $\mathrm{\ln k}$ vs $\frac{1}{T}$ is a straight line with slope equal to $-\frac{E_a}{R}$
Explanation:
(B) $\ln k=\ln A-\frac{E_{a}}{R T}$
$ \frac{1}{\mathrm{k}} \cdot \frac{\mathrm{dk}}{\mathrm{dT}}=\frac{+\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}^{2}} $
$\mathrm{E}_{\mathrm{a}} \uparrow$ temp. coefficient $\uparrow$
(C)
Option (C ) is wrong. $\Delta$k may be greater or lesser depending on temperature.
(D) $\ln k=\ln A-\frac{E_{a}}{R T}$
Slope of $\ln k$ vs $\frac{1}{T}$ is $\left(\frac{-E_{a}}{R}\right)$
For a chemical reaction $\mathrm{A}+\mathrm{B} \rightarrow$ Product, the order is 1 with respect to $\mathrm{A}$ and $\mathrm{B}$.
| $\mathrm{Rate}$ $\mathrm{mol~L^{-1}~S^{-1}}$ |
$\mathrm{[A]}$ $\mathrm{mol~L^{-1}}$ |
$\mathrm{[B]}$ $\mathrm{mol~L^{-1}}$ |
|---|---|---|
| 0.10 | 20 | 0.5 |
| 0.40 | $x$ | 0.5 |
| 0.80 | 40 | $y$ |
What is the value of $x$ and $y$ ?
The correct reaction profile diagram for a positive catalyst reaction.
Consider the following reaction that goes from A to B in three steps as shown below:
Choose the correct option
A student has studied the decomposition of a gas AB$_3$ at 25$^\circ$C. He obtained the following data.
| p (mm Hg) | 50 | 100 | 200 | 400 |
|---|---|---|---|---|
| relative t$_{1/2}$ (s) | 4 | 2 | 1 | 0.5 |
The order of the reaction is
Assuming $1 \,\mu \mathrm{g}$ of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ______ $\times\, 10^{-1} \mu \mathrm{g}$.
[Given : ln 10 = 2.303; log 2 = 0.30]
Explanation:
$ \begin{aligned} &\Rightarrow100=\left(\frac{30}{\ln 2}\right)\left[\ln \left(\frac{1}{w}\right)\right] \\ &\Rightarrow{\left[\frac{100 \times \log 2}{30}\right]=\log \left(\frac{1}{w}\right)} \\ &\Rightarrow1=\log \left(\frac{1}{w}\right) \\ &\Rightarrow\frac{1}{w}=10 \\ &\text { So } w=0.1 \mu g \end{aligned} $
The reaction between X and Y is first order with respect to X and zero order with respect to Y.
| Experiment | ${{[X]} \over {mol\,{L^{ - 1}}}}$ | ${{[Y]} \over {mol\,{L^{ - 1}}}}$ | ${{Initial\,rate} \over {mol\,{L^{ - 1}}\,{{\min }^{ - 1}}}}$ |
|---|---|---|---|
| I | 0.1 | 0.1 | $2 \times {10^{ - 3}}$ |
| I | L | 0.2 | $4 \times {10^{ - 3}}$ |
| III | 0.4 | 0.4 | $M \times {10^{ - 3}}$ |
| IV | 0.1 | 0.2 | $2 \times {10^{ - 3}}$ |
Examine the data of table and calculate ratio of numerical values of M and L. (Nearest Integer)
Explanation:
Using I & II
$\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\left(\frac{L}{0.1}\right) \quad \Rightarrow \quad \mathrm{L}=0.2$
Using I & III
$\frac{M \times 10^{-3}}{2 \times 10^{-3}}=\frac{0.4}{0.1} \quad \Rightarrow \quad \mathrm{M}=8$
$\frac{M}{L}=\frac{8}{0.2}=40$
For a reaction, given below is the graph of $\ln k$ vs ${1 \over T}$. The activation energy for the reaction is equal to ____________ $\mathrm{cal} \,\mathrm{mol}^{-1}$. (nearest integer)
(Given : $\mathrm{R}=2 \,\mathrm{cal} \,\mathrm{K}^{-1} \,\mathrm{~mol}^{-1}$ )
Explanation:
For the given first order reaction
$\mathrm{A} \rightarrow \mathrm{B}$
the half life of the reaction is $0.3010 \mathrm{~min}$. The ratio of the initial concentration of reactant to the concentration of reactant at time $2.0 \mathrm{~min}$ will be equal to ___________. (Nearest integer)
Explanation:
$ \begin{aligned} &K=\frac{0.693}{0.3010} \\\\ &K=2.30 \end{aligned} $
$\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left(\mathrm{A}_{0}\right)}{\left(\mathrm{A}_{\mathrm{t}}\right)}$
$A_{0} \rightarrow$ initial concentration of reactant
$A_{t} \rightarrow$ concentration of reactant at time $t$
$2.303=\frac{2.303}{2} \log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$2=\log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$100=\frac{A_{0}}{A_{t}}$
$\matrix{ {[A]} & \to & {[B]} \cr {{\mathop{\rm Reactant}\nolimits} } & {} & {{\mathop{\rm Product}\nolimits} } \cr } $
If formation of compound $[\mathrm{B}]$ follows the first order of kinetics and after 70 minutes the concentration of $[\mathrm{A}]$ was found to be half of its initial concentration. Then the rate constant of the reaction is $x \times 10^{-6} \mathrm{~s}^{-1}$. The value of $x$ is ______________. (Nearest Integer)
Explanation:
$=\frac{6930}{7 \times 6} \times 10^{-6}$
$=165 \times 10^{-6} \mathrm{~s}^{-1}$
$2 \mathrm{NO}+2 \mathrm{H}_{2} \rightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
The above reaction has been studied at $800^{\circ} \mathrm{C}$. The related data are given in the table below
| Reaction serial number | Initial Pressure of ${H_2}/kPa$ | Initial Pressure of $NO/kPa$ | Initial rate $\left( {{{ - dp} \over {dt}}} \right)/(kPa/s)$ |
|---|---|---|---|
| 1 | 65.6 | 40.0 | 0.135 |
| 2 | 65.6 | 20.1 | 0.033 |
| 3 | 38.6 | 65.6 | 0.214 |
| 4 | 19.2 | 65.6 | 0.106 |
The order of the reaction with respect to NO is ___________.
Explanation:
$ \mathrm{r}=\mathrm{K}[\mathrm{NO}]^{n}\left[\mathrm{H}_{2}\right]^{\mathrm{m}} $
From $1^{\text {st }}$ data
$ 0.135=\mathrm{K}[40]^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(1) $
From $2^{\text {nd }}$ data
$ 0.033=\mathrm{K}(20.1)^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(2) $
On dividing equation (1) by equation (2)
$ \begin{aligned} &\frac{0.135}{0.033}=\left(\frac{40}{20.1}\right)^{n} \\\\ &4=(2)^{n} \\\\ &\therefore n=2 \\\\ &\therefore \text { Order of reaction w.r.t. NO is } 2 . \end{aligned} $
For a reaction $\mathrm{A} \rightarrow 2 \mathrm{~B}+\mathrm{C}$ the half lives are $100 \mathrm{~s}$ and $50 \mathrm{~s}$ when the concentration of reactant $\mathrm{A}$ is $0.5$ and $1.0 \mathrm{~mol} \mathrm{~L}^{-1}$ respectively. The order of the reaction is ______________ . (Nearest Integer)
Explanation:
$ \begin{array}{ll} \mathrm{t}_{1 / 2}=100 \,\mathrm{sec} & \mathrm{a}_{0}=0.5 \\ \mathrm{t}_{1 / 2}=50 \,\mathrm{sec} & \mathrm{a}_{0}=1 \end{array} $
$\frac{100}{50}=\left(\frac{1}{0 \cdot 5}\right)^{n-1}$
$(2)=(2)^{n-1}$
$\mathrm{n}-1=1$
$n=2$
For the decomposition of azomethane.
CH3N2CH3(g) $\to$ CH3CH3(g) + N2(g), a first order reaction, the variation in partial pressure with time at 600 K is given as
The half life of the reaction is __________ $\times$ 10$-$5 s. [Nearest integer]
Explanation:
$ \begin{aligned} &\ln A=\ln A_{0}-k t \\ &\text { Hence Slope }=-k \\ &-k=-3.465 \times 10^{4} \\ &k=\frac{0.693}{t_{1 / 2}} \\ &3.465 \times 10^{4}=\frac{0.693}{t_{1 / 2}} \\ &t_{1 / 2}=2 \times 10^{-5} \mathrm{~s} \end{aligned} $
The half life for the decomposition of gaseous compound $\mathrm{A}$ is $240 \mathrm{~s}$ when the gaseous pressure was 500 Torr initially. When the pressure was 250 Torr, the half life was found to be $4.0$ min. The order of the reaction is ______________. (Nearest integer)
Explanation:
$\left(t_{1 / 2}\right)_{A}=4 \min =4 \times 60=240$ sec when $P=250$ torr
If means half-life is independent of concentration of reactant present.
$\therefore$ Order of reaction $=1$
For the reaction P $\to$ B, the values of frequency factor A and activation energy EA are 4 $\times$ 1013 s$-$1 and 8.3 kJ mol$-$1 respectively. If the reaction is of first order, the temperature at which the rate constant is 2 $\times$ 10$-$6 s$-$1 is _____________ $\times$ 10$-$1 K.
(Given : ln 10 = 2.3, R = 8.3 J K$-$1 mol$-$1, log2 = 0.30)
Explanation:
The equation
k = (6.5 $\times$ 1012s$-$1)e$-$26000K/T
is followed for the decomposition of compound A. The activation energy for the reaction is ________ kJ mol$-$1. [nearest integer]
(Given : R = 8.314 J K$-$1 mol$-$1)
Explanation:
$ \begin{aligned} \frac{E_{a}}{R T} &=\frac{26000}{T} \\\\ E_{a} &=26000 \times 8.314 \\\\ &=216164 \mathrm{~J} \\\\ &=216 \mathrm{~kJ} \end{aligned} $
The activation energy of one of the reactions in a biochemical process is 532611 J mol$-$1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x $\times$ 10$-$3 k310. The value of x is ____________.
[Given : $\ln 10 = 2.3$, R = 8.3 J K$-$1 mol$-$1]
Explanation:
where $\mathrm{K}_2$ is at $310 \mathrm{~K} \,\& \mathrm{~K}_1$ is at $300 \mathrm{~K}$
$ \begin{aligned} & \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=6.9 \\\\ & =3 \times \ln 10 \\\\ & \ln \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\ln 10^3 \\\\ & \mathrm{~K}_2=\mathrm{K}_1 \times 10^3 \\\\ & \mathrm{~K}_1=\mathrm{K}_2 \times 10^3 \end{aligned} $
So K = 1
A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ___________. (Nearest integer)
(Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)
Explanation:
$ \begin{aligned} &\frac{0.693}{200}=\frac{2.303}{83} \log \frac{A_{0}}{A} \\\\ &\frac{A}{A_{0}}=0.75 \end{aligned} $
Hence, percentage of original activity remaining after 83 days is $75 \%$
For a first order reaction A $\to$ B, the rate constant, k = 5.5 $\times$ 10$-$14 s$-$1. The time required for 67% completion of reaction is x $\times$ 10$-$1 times the half life of reaction. The value of x is _____________ (Nearest integer)
(Given : log 3 = 0.4771)
Explanation:
Nearest integer $=16$
It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ____________ kJ mol$-$1. [nearest integer]
(Given ln10 = 2.3, R = 8.3 J K$-$1 mol$-$1, log 2 = 0.30)
Explanation:
(Rate constant)
$\mathrm{K}_{2}=2 \mathrm{~K}_{1}$, on increase temperature by $9 \mathrm{~K}$
$\mathrm{T}_{2}=309 \mathrm{~K}$
$\mathrm{Ea}=?$
$\log \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{2.3 \mathrm{R}}\left[\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{2} \cdot \mathrm{T}_{1}}\right]$
$\log 2=\frac{\mathrm{Ea}}{2.3 \times 8.3}\left[\frac{9}{309 \times 300}\right]$
$\mathrm{Ea}=\frac{0.3 \times 309 \times 300 \times 2.3 \times 8.3}{9}$
$=58988.1 \mathrm{~J} / \mathrm{mole}$
$\simeq 59 \mathrm{~kJ} / \mathrm{mole}$
The rate constant for a first order reaction is given by the following equation:
$\ln k = 33.24 - {{2.0 \times {{10}^4}\,K} \over T}$
The activation energy for the reaction is given by ____________ kJ mol$-$1. (In nearest integer) (Given : R = 8.3 J K$-$1 mol$-$1)
Explanation:
Given: $\ln k=33.24-\frac{2.0 \times 10^4}{\mathrm{~T}}$
$ \begin{aligned} &\therefore \text { on comparing } \frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{R}}=2.0 \times 10^4 \\\\ &\therefore \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times \mathrm{R} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times 8.3 \mathrm{~J} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=16.6 \times 10^4 \mathrm{~J}=166 \mathrm{~kJ} \end{aligned} $
Catalyst A reduces the activation energy for a reaction by 10 kJ mol$-$1 at 300 K. The ratio of rate constants, ${{{}^kT,\,Catalysed} \over {{}^kT,\,Uncatalysed}}$ is ex. The value of x is ___________. [nearest integer]
[Assume that the pre-exponential factor is same in both the cases. Given R = 8.31 J K$-$1 mol$-$1]
Explanation:
A catalyst increases the rate of a reaction by lowering the activation energy so that more reactant molecules collide with enough energy to surmount the smaller energy barrier.
In the Question,
Given : $\mathrm{E}_{\text {cat }}-\mathrm{E}_{\text {uncat }}=10 \mathrm{~kJ} / \mathrm{mol}$
$ \mathrm{T}=300 \mathrm{~K} $
According to Arrhenius Equation,
$ \begin{aligned} & \mathrm{K}=A \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \\\\ & \frac{\mathrm{E}_{\text {cat }}}{\mathrm{E}_{\text {uncat }}}=e^{\frac{E a-E_a^1}{R T}} \end{aligned} $
$ \begin{aligned} & =e^{\frac{10 \times 10^3}{8.21 \times 300}} \\\\ & =e^4 \\\\ & =4 \end{aligned} $