Chemical Kinetics and Nuclear Chemistry
Consider a reaction $A+R \rightarrow$ Product. The rate of this reaction is measured to be $k[A][R]$. At the start of the reaction, the concentration of $R,[R]_0$, is 10-times the concentration of $A,[A]_0$. The reaction can be considered to be a pseudo first order reaction with assumption that $k[R]=k^{\prime}$ is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is $40 \%$ complete, is ___________.
[ $k$ and $k^{\prime}$ represent corresponding rate constants]
Explanation:
$\begin{aligned} & \text { Rate }=k[\mathrm{~A}][\mathrm{R}] \\\\ & \operatorname{Rate}_1=k\left(0.6 \mathrm{~A}_0\right) \times 9.6 \mathrm{~A}_0\end{aligned}$
$\begin{aligned} & \text { Rate }=\mathrm{k}^{\prime}[\mathrm{A}], \mathrm{k}^{\prime}=\mathrm{k}[\mathrm{R}] \end{aligned}$
The relative error between the actual rate and the pseudo-first order rate is computed by:
$ \text{Relative Error (%)} = \frac{\text{Rate}_2 - \text{Rate}_1}{\text{Rate}_1} \times 100 $
Plugging the calculations:
$ \text{Relative Error (%)} = \frac{(0.6 \times 10 - 0.6 \times 9.6)}{0.6 \times 9.6} \times 100 = 4.1666\% $
A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into $\mathrm{Pb}-206$ while the rest of it remains undisintegrated.
When the age of the sample is $\mathbf{P} \times 10^8$ years, the ratio of mass of $\mathrm{Pb}-206$ to that of $\mathrm{U}-238$ in the sample is found to be 7. The value of $\mathbf{P}$ is _______.
[Given: Half-life of $\mathrm{U}-238$ is $4.5 \times 10^9$ years; $\log _e 2=0.693$ ]
Explanation:
- Given Data:
- Ratio of mass of $ \mathrm{Pb}^{206} $ to $ \mathrm{U}^{238} $ = $ \frac{7}{1} $
- Mass of $ \mathrm{Pb}^{206} $ = 7 g
- Mass of $ \mathrm{U}^{238} $ = 1 g
- Half-life of $ \mathrm{U}^{238} $ = $ 4.5 \times 10^9 $ years
- $ \log_e 2 = 0.693 $
- Calculate Moles:
- Moles of $ \mathrm{Pb}^{206} $:
$ \text{Moles of } \mathrm{Pb}^{206} = \frac{7}{206} = 34 \times 10^{-3} \text{ moles} $
- Moles of $ \mathrm{U}^{238} $:
$ \text{Moles of } \mathrm{U}^{238} = \frac{1}{238} = 4.2 \times 10^{-3} \text{ moles} $
- Initial Moles of $ \mathrm{U}^{238} $ ( $ N_0 $ ):
$ N_0 = \text{Moles of } \mathrm{U}^{238} \text{ initially} = \text{Moles of } \mathrm{Pb}^{206} + \text{Moles of remaining } \mathrm{U}^{238} $
$ N_0 = 34 \times 10^{-3} + 4.2 \times 10^{-3} = 38.2 \times 10^{-3} \text{ moles} $
- Calculate the Time:
- Using the decay formula:
$ \lambda t = \ln \frac{N_0}{N_t} $
- Where $ \lambda $ (decay constant):
$ \lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{4.5 \times 10^9 \text{ years}} $
- Calculate $ \ln \frac{N_0}{N_t} $:
$ \ln \frac{N_0}{N_t} = \ln \frac{38.2}{4.2} = \ln 9.09 $
- Simplify $ \ln 9.09 $:
$ \ln 9.09 \approx 2.21 $
- Plug in the Values:
$ t = \frac{4.5 \times 10^9 \text{ years}}{0.693} \times \ln 9.09 $
$ t = 4.5 \times 10^9 \text{ years} \times \frac{2.21}{0.693} $
$ t \approx 4.5 \times 10^9 \times 3.184 \approx 14.33 \times 10^9 \text{ years} $
- Convert to $ P \times 10^8 $:
$ t = P \times 10^8 \text{ years} $
$ P \times 10^8 = 14.33 \times 10^9 $
$ P = 143.28 $
Thus, the value of $ P $ is $ \boxed{143.28} $.
Consider the following reaction,
$ 2 \mathrm{H}_2(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) $
which follows the mechanism given below :
$ \begin{array}{ll} 2 \mathrm{NO}(\mathrm{g}) \stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) & \text { (fast equlibrium) } \\\\ \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_2} \mathrm{~N}_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \text { (slow reaction) } \\\\ \mathrm{N}_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_3} \mathrm{~N}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \text { (fast reaction) } \end{array} $
The order of the reaction is __________.
Explanation:
The given reaction is:
$2 \mathrm{H}_2(\mathrm{~g}) + 2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{~g}) + 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
And it follows the mechanism below:
$ \begin{array}{ll} 2 \mathrm{NO}(\mathrm{g}) \stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) & \text { (fast equilibrum) } \\\\ \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) + \mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_2} \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \text { (slow reaction) } \\\\ \mathrm{N}_2 \mathrm{O}(\mathrm{g}) + \mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_3} \mathrm{~N}_2(\mathrm{~g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \text { (fast reaction) } \end{array} $
To determine the order of the reaction, we need to focus on the slow step, as it is the rate-determining step. The slow step is:
$ \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) + \mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_2} \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) $
The rate law for this step can be written as:
$ \text{Rate} = k_2 [\mathrm{N}_2\mathrm{O}_2][\mathrm{H}_2] $
However, $[\mathrm{N}_2 \mathrm{O}_2]$ is an intermediate and its concentration can be expressed in terms of the reactants. From the fast equilibrium step:
$ 2 \mathrm{NO}(\mathrm{g}) \stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) $
The equilibrium constant for this step can be written as:
$\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}$ = $ K = \frac{[\mathrm{N}_2 \mathrm{O}_2]}{[\mathrm{NO}]^2} $
Hence, the concentration of the intermediate $\mathrm{N}_2 \mathrm{O}_2$ can be expressed as:
$ [\mathrm{N}_2 \mathrm{O}_2] = K[\mathrm{NO}]^2 $
Substituting this into the rate law for the slow step gives:
$ \text{Rate} = k_2 K [\mathrm{NO}]^2[\mathrm{H}_2] $
Let's combine $k_2$ and $K$ into a single constant, say $k'$:
$ \text{Rate} = k' [\mathrm{NO}]^2[\mathrm{H}_2] $
This implies that the reaction is second-order in $\mathrm{NO}$ and first-order in $\mathrm{H}_2$. Therefore, the overall order of the reaction is:
$ 2 + 1 = 3 $
Thus, the order of the reaction is 3.
Explanation:
Number of moles of $_{92}^{238}U$ present initially
= ${{68 \times {{10}^{ - 6}}} \over {238}}$
After three half-lifes, moles of $_{92}^{238}U$ decayed
$ = {{68 \times {{10}^{ - 6}}} \over {238}} \times \left( {1 - {1 \over {{2^3}}}} \right)$
$ = {{68 \times {{10}^{ - 6}}} \over {238}} \times {7 \over 8}$
Therefore, number of $\alpha $-particles emitted
$ = {{68 \times {{10}^{ - 6}}} \over {238}} \times {7 \over 8} \times 8 \times 6.023 \times {10^{23}}$
$ = 1.204 \times {10^{18}}$
$ \approx 1.2 \times {10^{18}}$
Thus, the correct answer is 1.2.
$2{N_2}{O_5}(g)\buildrel \Delta \over \longrightarrow 2{N_2}{O_4}(g) + {O_2}(g)$
is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y $ \times $ 103 s, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5 $ \times $ 10-4s-1, assuming ideal gas behaviour, the value of Y is ...............
Explanation:
At initial t = 0 and final t = y $ \times $ 103 sec
$\matrix{ {2{N_2}{O_5}(g)\buildrel \Delta \over \longrightarrow } & {2{N_2}{O_4}(g) + } & {{O_2}(g)} \cr 1 & 0 & 0 \cr {1 - 2p} & {2p} & p \cr } $
pTotal = 1 $-$ 2p + 2p + p
1.4 = 1 + p
p = 0.45 atm
According to first order reaction,
$k = {{2.303} \over t}\log {{{p_i}} \over {{p_i} - 2p}}$
pi = 1 atm (given)
2p = 2 $ \times $ 0.45 = 0.9 atm
On substituting the values in above equation,
$2k.t = 2.303\log {1 \over {1 - 0.9}}$
$2 \times 5 \times {10^{ - 4}} \times y \times {10^3} = 2.303\log {1 \over {0.1}}$
$y = 2.303 = 2.3$
Note : Unit of rate constant (k), i.e. s$-$1 represents that it is a first order reaction.
The rate of the reaction for [A] = 0.15 mol dm-3, [B] = 0.25 mol dm-3 and [C] = 0.15 mol dm-3 is found to be Y $ \times $ 10-5 mol dm-3s-1. The value of Y is .................
Explanation:
${{{{(Rate)}_1}} \over {{{(Rate)}_2}}} = {{{{[0.2]}^x}{{[0.1]}^y}{{[0.1]}^z}} \over {{{[0.2]}^x}{{[0.2]}^y}{{[0.1]}^z}}} = {{6 \times {{10}^{ - 5}}} \over {6 \times {{10}^{ - 5}}}}$
$ \Rightarrow $ y = 0
${{{{(Rate)}_1}} \over {{{(Rate)}_3}}} = {{{{[0.2]}^x}{{[0.1]}^y}{{[0.1]}^z}} \over {{{[0.2]}^x}{{[0.1]}^y}{{[0.2]}^z}}} = {{6 \times {{10}^{ - 5}}} \over {1.2 \times {{10}^{ - 4}}}}$
$ \Rightarrow $ z = 1
${{{{(Rate)}_1}} \over {{{(Rate)}_4}}} = {{{{[0.2]}^x}{{[0.1]}^y}{{[0.1]}^z}} \over {{{[0.3]}^x}{{[0.1]}^y}{{[0.1]}^z}}} = {{6 \times {{10}^{ - 5}}} \over {9 \times {{10}^{ - 5}}}}$
$ \Rightarrow $ x = 1
So, rate = k[A]1[C]1
From exp-Ist,
Rate = $6.0 \times {10^{ - 5}}$ mol dm$ - $3 s$ - $1
6.0 $ \times $ 10$ - $5 = k[0.2]1[0.1]1
k = 3 $ \times $ 10$ - $3
Given, [A] = 0.15 mol dm$ - $3
[B] = 0.25 mol dm$ - $3
[C] = 0.15 mol dm$ - $3
$ \therefore $ Rate = (3 $ \times $ 10$ - $3) $ \times $ [0.15]1[0.25]0[0.15]1
= 3 $ \times $ 10$ - $3 $ \times $ 0.15 $ \times $ 0.15
Rate = 6.75 $ \times $ 10$ - $5 mol dm$ - $3 s$ - $1
Thus, Y = 6.75
The activation energy of the backward reaction exceeds that of the forward reaction by $2RT$ (in $J\,mo{l^{ - 1}}$). If the pre-exponential factor of the forward reaction is $4$ times that of the reverse reaction, the absolute value of $\Delta {G^ \circ }$ (in $J\,mo{l^{ - 1}}$ ) for the reaction at $300$ $K$ is ____________.
(Given; $\ln \left( 2 \right) = 0.7,RT = 2500$ $J\,mo{l^{ - 1}}$ at $300$ $K$ and $G$ is the Gibbs energy)
Explanation:
For the reversible reaction,
$ \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{AB}(g) $
(i) Pre-exponential factor for forward reaction $\left(A_f\right)=4 \times$ pre-exponential factor for backward reverse reaction $\left(A_b\right)$
$ A_f=4 \times A_b $
(ii) Activation energy for backward reaction $\left(E_b\right)_f-$ Activation energy $=2 \mathrm{RT}$ for forward reaction $\left(\mathrm{E}_a\right)_{f^{-}}$
$ \mathrm{A}_b-\mathrm{A}_f=2 \mathrm{RT} $
$ \begin{aligned} & \text {(iii) Equilibrium constant }\left(\mathrm{K}_{e q}\right)=\frac{\mathrm{K}_f}{\mathrm{~K}_b} \\\\ & \mathrm{~K}_{e q}=\frac{\mathrm{A}_f \times e^{-\left(\mathrm{E}_a\right)_f} / \mathrm{RT}}{\mathrm{A}_b \times e^{-\left(\mathrm{E}_a\right)_b} / \mathrm{RT}} \end{aligned} $
$ \begin{aligned} \mathrm{K}_{e q} & =\frac{\mathrm{A}_f}{\mathrm{~A}_b} \times e^{\left[-\left(\mathrm{E}_a\right)_f-\left(\mathrm{E}_a\right)_b\right]} / \mathrm{RT} \\\\ \mathrm{K}_{e q} & =\frac{4 \mathrm{~A}_b}{\mathrm{~A}_b} \times e^{-[2 \mathrm{RT} / \mathrm{RT}]} \\\\ \mathrm{K}_{e q} & =4 \times e^{-2} \\\\ \ln \mathrm{K}_{e q} & =\ln 4+2 \ln \\\\ \ln \mathrm{K}_{e q} & =2 \ln 2+2=2 \times 0.7+2 \\\\ & =(1.4+2)=3.4 \end{aligned} $
The expression for the change in Gibbs free energy of reaction $(\Delta \mathrm{G})$ is given as:
$ \Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \ln k $
At equilibrium, $\Delta \mathrm{G}=0$
$ \Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}_{e q} $
Substituting the value of RT and $\ln \mathrm{K}_{e q}$
$ \begin{aligned} \Delta \mathrm{G}^{\circ} & =-2500 \mathrm{~J} \mathrm{~mol}^{-1} \times 3.4 \\\\ \Delta \mathrm{G}^{\circ} & =-8500 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} $
The absolute value of Gibbs free energy $\left(\Delta \mathrm{G}^{\circ}\right)$ for the reaction at $300 \mathrm{~K}$ is $8500 \mathrm{~J} \mathrm{~mol}^{-1}$.
Explanation:
${}_{92}{U^{238}} \to {}_{82}P{b^{206}} + 8\,{}_2H{e^4}(g) + 6{}_{ - 1}{\beta ^0}$
To calculate pressure, only gaseous products need to be considered.
Initially, only 1 mol of air is present and finally, after complete decay, 8 moles of $_2^4He$ gas are produced and 1 mol of air will also remain in the mixture.
Ratio of the final pressure to the initial pressure $ = {{8 + 1} \over 1} = 9$
Explanation:
In complex, $\mathop {{{[Fe{{({C_2}{O_4})}_2}{{({H_2}O)}_2}]}^{2 - }}}\limits_{Diaquodioxalatoferrate\,(II)} $,
Fe is in +2 oxidation state.
In acidic medium, $KMn{O_4}$ oxidises $F{e^{2 + }}$ to $F{e^{3 + }}$,
$2MnO_4^ - + 16{H^ + } + 10F{e^{2 + }} \to 2M{n^{2 + }} + 8{H_2}O + 10F{e^{3 + }}$
or $MnO_4^ - + 8{H^ + } + 5F{e^{2 + }} \to M{n^{2 + }} + 4{H_2}O + 5F{e^{3 + }}$
${{Rate\,of\,change\,of\,[{H^ + }]} \over {Rate\,of\,change\,of\,[MnO_4^ - ]}} = {8 \over 1} = 8$
Explanation:
$ t=\frac{2.303}{k} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]} $
When the compound is decomposed to $1 / 8$ th of its initial value then the time taken is
$ t_{1 / 8}=\left(\frac{2.303}{k}\right) \log \frac{1}{(1 / 8)}=\left(\frac{2.303}{k}\right) \log 8 $ .........(1)
When the compound is decomposed to $1 / 10$ th of its initial value then the time taken is
$ t_{1 / 10}=\left(\frac{2.303}{k}\right) \log \frac{1}{(1 / 10)}=\left(\frac{2.303}{k}\right) \log 10 $ ........(2)
Dividing Eq. (1) by Eq. (2), we get
$ \frac{t_{1 / 8}}{t_{1 / 10}}=\frac{\log 8}{\log 10}=\log \left(2^3\right)=3 \times 0.3=0.9 $
So, the value of
$ \frac{\left[t_{1 / 8}\right]}{\left[t_{1 / 10}\right]} \times 10=9 $
${}_{29}^{63}Cu$ + ${}_1^1H$ $\to$ $6{}_0^1n$ + ${}_2^4\alpha $ + 2${}_1^1H$ + X
Explanation:
$ { }_{29}^{63} \mathrm{Cu}+{ }_1^1 \mathrm{H} \rightarrow 6{ }_0^1 n+{ }_2^4 \mathrm{He}+2{ }_1^1 \mathrm{H}+\mathrm{X} $
Equating mass numbers on both the sides, we get
$ 63+1=1 \times 6+4 \times 1+1 \times 2 + X$
$ \Rightarrow $ $X=64-12=52 $
Equating atomic numbers on both the sides, we get
$ 29+1=6 \times 0+2+2 \times 1+\mathrm{Y} $
$\Rightarrow \mathrm{Y}=30-4=26 $
So, the element is ${ }_{26}^{52} \mathrm{Fe}$ and iron is a $d$-block element, which belongs to Group 8 of the periodic table.
Explanation:
To determine the number of neutrons emitted during the nuclear fission of ${}_{92}^{235}U$ resulting in the products ${}_{54}^{142}Xe$ and ${}_{38}^{90}Sr$, we need to ensure the conservation of mass number and atomic number.
The mass number (A) and atomic number (Z) have to be conserved. This means that the sum of the mass numbers and atomic numbers of the products (including any neutrons emitted) must equal those of the uranium nucleus undergoing fission.
Let's start by writing down the conservation of mass number and atomic number:
- Conservation of Mass Number:
${235 = 142 + 90 + n \times 1}$
Here, $n$ represents the number of neutrons released. We can now calculate $n$:
$ n = 235 - (142 + 90) = 235 - 232 = 3 $
- Conservation of Atomic Number:
${92 = 54 + 38 + 0 \times n}$
Neutrons do not contribute to the atomic number as they have no charge.
This calculation shows that three neutrons are needed to satisfy the conservation of mass number. The atomic number conservation also coincides, as neutrons do not alter it. Therefore, the number of neutrons emitted during this fission process is $3$.
| [R] molar | 1.0 | 0.75 | 0.40 | 0.10 |
|---|---|---|---|---|
| t (min.) | 0.0 | 0.05 | 0.12 | 0.18 |
The total number of $\alpha$ and $\beta$ particles emitted in the nuclear reaction $_{92}^{238}U \to _{82}^{214}Pb$ is _________.
Explanation:
$ { }_{92}^{238} \mathrm{U} \longrightarrow{ }_{82}^{214} \mathrm{~Pb} $
It is also written as
So, to find the alpha particle you can solve it by atomic mass of $\mathrm{Pb}$
$ =\text { Atomic mass of } \mathrm{U}-4 a $
Atomic mass of $U=238$
Atomic mass of $\mathrm{Pb}=214$
So, by putting the values in equation (i), you will get
$ 206=238-4 a $
$ \begin{aligned} \therefore \alpha & =\frac{238-214}{4} =6 \end{aligned} $
Now, find the beta particle by using formula
Atomic number of $\mathrm{U}=$ Atomic number of $\mathrm{Pb}+2 \alpha+\beta$.
Where, atomic no. of $U$ is 92 and atomic no. of $\mathrm{Pb}$ is 82 and the value of $\alpha$ that we have find is 6. By putting the values you will get.
$ 92=82+2 \times 6=\beta $
Therefore, $ \beta=92-94=2$
Thus, the number of alpha and beta particles is 6 and 2.
Match the rate expressions in LIST-I for the decomposition of $X$ with the corresponding profiles provided in LIST-II. $X_{\mathrm{s}}$ and $\mathrm{k}$ are constants having appropriate units.
| List-I | List-II |
|---|---|
| (I) rate $=\frac{\mathrm{k}[\mathrm{X}]}{\mathrm{X}_{\mathrm{s}}+[\mathrm{X}]}$ under all possible initial concentrations of $\mathrm{X}$ |
(P)  |
| (II) rate $=\frac{k[X]}{X_{s}+[X]}$ where initial concentrations of $X$ are much less than $X_{s}$ |
(Q)  |
| (III) rate $=\frac{k[X]}{X_{s}+[X]}$ where initial concentrations of $\mathrm{X}$ are much higher than $X_{s}$ |
(R)  |
| (IV) rate $=\frac{k[X]^{2}}{X_{s}+[X]}$ where initial concentration of $X$ is much higher than $\mathrm{X}_{\mathrm{s}}$ |
(S)  |
(T)  |
The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is due to
In the reaction, P + Q $\to$ R + S, the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is
Bombardment of aluminium by $\alpha$-particle leads to its artificial disintegration in two ways : (i) and (ii) as shown. Products X, Y and Z, respectively, are
Plots showing the variation of the rate constant ($k$) with temperature ($T$) are given below. The point that follows Arrhenius equation is
For a first-order reaction A $\to$ P, the temperature (T) dependent rate constant (k) was found to follow the equation $\log k = - (2000){1 \over T} + 6.0$. The pre-exponential factor A and activation energy $E_a$, respectively, are
Under the same reaction conditions, initial concentration of 1.386 mol dm$^{-3}$ of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio $\left( {{{{k_1}} \over {{k_0}}}} \right)$ of the rate constants for first order ($k_1$) and zero order ($k_0$) of the reactions is:
Which of the following option is correct?
In living organisms, circulation of ${ }^{14} \mathrm{C}$ from atmosphere is high, so the carbon content is constant in organism.
Carbon dating can be used to find out the age of earth crust and rocks.
Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay; hence, the carbon content remains constant in living organism.
Carbon dating cannot be used to determine concentration of ${ }^{14} \mathrm{C}$ in dead beings.
What should be the age of fossil for meaningful determination of its age?
6 years
6000 years
60,000 years
It can be used to calculate any age
A nuclear explosion has taken place leading to increase in concentration of ${ }^{14} \mathrm{C}$ in nearby areas. ${ }^{14} \mathrm{C}$ concentration is $\mathrm{C}_1$ in nearby areas and $C_2$ in areas far away. If the age of the fossil is determined to be $T_1$ and $T_2$ at the places respectively then,
The age of the fossil will increase at the place where explosion has taken place and $\mathrm{T}_1-\mathrm{T}_2=\frac{1}{\lambda} \ln \frac{\mathrm{C}_1}{\mathrm{C}_2}$.
The age of the fossil will decrease at the place where explosion has taken place and $\mathrm{T}_1-\mathrm{T}_2=\frac{1}{\lambda} \ln \frac{\mathrm{C}_1}{\mathrm{C}_2}$.
The age of fossil will be determined to be the same.
$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\mathrm{C}_1}{\mathrm{C}_2}$.
$ \text { Match Column I with Column II : } $
| Column I | Column II | ||
|---|---|---|---|
| (A) | $\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CD}_3$ on treatment with alc. KOH gives $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CD}_3$ as a major product. | (P) | E1 reaction |
| (B) | $ \begin{aligned} &\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CH}_3\\ &\text { reacts faster than }\\ &\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CD}_3 . \end{aligned} $ |
(Q) | E2 reaction |
| (C) | $ \mathrm{Ph}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br} $ on treatment with $ \mathrm{C}_2 \mathrm{H}_5 \mathrm{OD} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-} $ gives $\mathrm{Ph}-\mathrm{CD}=\mathrm{CH}_2$ as the major product. |
(R) | E1 cb reaction |
| (D) | $\mathrm{PhCH}_2, \mathrm{CH}_2 \mathrm{Br}$ and $\mathrm{PhCD}_2 \mathrm{CH}_2 \mathrm{Br}$ react with same rate. | (S) | First order reaction |
$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{Q}) ; \mathrm{C} \rightarrow( \mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{P}, \mathrm{~S})] .$
$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{Q}) ; \mathrm{C} \rightarrow(\mathrm{R}, \mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{P})] .$
$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{Q}) ; \mathrm{C} \rightarrow(\mathrm{R}, \mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{P}, \mathrm{~S})] .$
$ [\mathrm{A} \rightarrow(\mathrm{Q}, P) ; \mathrm{B} \rightarrow(\mathrm{Q}) ; \mathrm{C} \rightarrow(\mathrm{R}, \mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{P}, \mathrm{~S})] .$
Note: The bond between C and deuterium is much strong than that of C and H , hence, elimination takes place from $\mathrm{C}-\mathrm{H}$ bonds only. This chemical reaction undergoes E2 mechanism as two substituents, H and Br atoms are removed simultaneously from the compound to form alkene.
$2X + Y\buildrel k \over \longrightarrow P$ the rate of reaction is ${{d[P]} \over {dt}} = k[X]$. Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are)
(Use : ln 2 = 0.693)
([P]0 is the initial concentration of P)
x1, x2, x3 and x4 are particles/radiation emitted by the respective isotopes. The correct option(s) is(are)
The % yield of ammonia as a function of time in the reaction
N2(g) + 3H2(g) $\rightleftharpoons$ 2NH3(g), $\Delta$H < 0 at (P, T1) is given below:
If this reactions is conducted at (P, T2), with T2 > T1, the % yield of ammonia as a function of time is represented by
9Be4 + X $\to$ 8Be4 + Y
(X, Y) is (are) :
2N2O5 (g) $\to$ 4NO2 (g) + O2 (g)
| Observation No. | Time (in minute) | Px (in mm of Hg) |
|---|---|---|
| 1 | 0 | 800 |
| 2 | 100 | 400 |
| 3 | 200 | 200 |
(i) What is the order of the reaction to X?
(ii) Find the rate constant
(iii) Find the time for 75% completion of the reaction.
(iv) Find the total pressure when pressure of X is 700 mm of Hg
Explanation:
The given reaction is : 2X(g) $\to$ 3Y(g) + 2Z(g)
(1) Partial pressure (of X) trend :
800 mm Hg $\buildrel {100\,\min } \over \longrightarrow $ 400 mm Hg $\buildrel {100\,\min } \over \longrightarrow $ 200 mm Hg Half-life is constant, hence, it is a first order reaction.
(2) For a first order reaction, rate constant $k = {{0.693} \over {half - life}}$
or, $k = {{0.693} \over {100}} = 6.93 \times {10^{ - 3}}$ min$-$1
(3) Time required for the completion of 75% of the reaction is equal to 2 half lives. Hence, Time taken = (2 $\times$ 100) = 200 min
(4) 
$\therefore$ Total pressure = (700 + 150 + 100) mm Hg = 950 mm Hg
Fill in the blanks:
(A) $_{92}^{235}$U + $_{0}^{1}$n $\to$ $_{52}^{137}$A + $_{40}^{97}$B + ____________.
(B) $_{34}^{82}$Se $\to$ 2 ${}_{ - 1}{e^0}$ + __________.
Explanation:
Subjective Answer
(A) Assume that, the element or component that should be present at the blank space in the given reaction is X:
$_{92}^{235}U + _0^1n \to _{52}^{137}A + _{40}^{97}B + \underline X $
The atomic number of uranium is 92 and atomic mass number is 235. The atomic number of element A and B is 52 and 40 respectively. The atomic mass number of elements A and B are 137 and 97 respectively.
The element with atomic number 52 and atomic mass number 137 is tellurium, Te. Therefore, element A is tellurium.
The element with atomic number 40 and atomic mass number 97 is zirconium, Zr. Therefore, element B is zirconium.
To find out X, we need to balance the atomic mass number and atomic number on both sides.
On the left side of the reaction, the sum of atomic number is 92 + 0 = 92 and that of atomic mass number is 235 + 1 = 236.
On the right side of the reaction, the sum of atomic number is 52+40 =92 and that of atomic mass number is 137+ 97 = 234.
The calculation shows that, the missing component at the blank space will have atomic number 0 and atomic mass number 2. The component with 0 atomic number will be neutron which will have atomic mass number 1. Thus, X equals to 2 neutrons.
Therefore, the complete equation with filled blank space can be written as,
$_{92}^{235}U + _0^1n \to _{52}^{137}A + _{40}^{97}B + \underline {2\,_0^1n} $
(B) Assume that, the element or component that should be present at the blank space in the given reaction is Y.
$_{34}^{82}Se \to 2_{ - 1}^0e + \underline Y $
The atomic number of selenium is 34 and atomic mass number is 82. The selenium is given 2 electrons and component Y.
To find out Y, we need to balance the atomic mass number and atomic number on both sides.
On the left side of the reaction, atomic number is 34 and that of atomic mass number is 92 as shown.
On the right side of the reaction, the electrons have atomic number as $–$1. There are two electrons, therefore, 2x ($–$1) = $–$2.
The atomic mass number of component Y must be 82.
The atomic number of component Y must be 36, as 36 + ($–$2) = 34 to balance both sides.
The element with atomic number 36 and atomic mass number 82 is Krypton and is denotes as Kr. Therefore, the complete equation with filled blank space can be written as,
$_{34}^{82}Se \to 2_{ - 1}^0e + _{36}^{82}\underline {Kr} $
Final Answer
(A) $_{92}^{235}U + _0^1n \to _{52}^{137}A + _{40}^{97}B + \underline {2\,_0^1n} $
(B) $_{34}^{82}Se \to 2_{ - 1}^0e + _{36}^{82}\underline {Kr} $
Hints:
The atomic number at the right side of equation should be equal to the sum of atomic number at the left side for both equations.
For the following reaction
2X(g) $\to$ 3Y(g) + 2Z(g)
assuming ideal gas conditions, the data for change of partial pressure with time is as follows:
$ \begin{array}{llll} \hline \text { Time (in min) } & 0 & 100 & 200 \\ \hline \begin{array}{l} \text { Partial pressure of X } \\ \text { (in mm of Hg) } \end{array} & 800 & 400 & 200 \end{array} $
Calculate
(A) Order of reaction.
(B) Rate constant.
(C) Time taken for 75% completion of reaction.
(D) Total pressure of the reaction mixture when $p_x=700$ mm.
Explanation:
The given chemical reaction is,
2X(g) $\to$ 3Y(g) + 2Z(g)
The data given for the above chemical reaction is,
| Time (in min) | 0 | 100 | 200 |
|---|---|---|---|
| Partial pressure of X (in mm of Hg) | 800 | 400 | 200 |
(A) From the given data, the duration of decreasing partial pressure of X from 800 to 400 and from 400 to 200, is 100 minutes for both. Therefore, the time required for decreasing of partial pressure is constant for decrement of pressure from 800 to 400 and 400 to 200.
Thus, the above chemical reaction is first order reaction.
(B) For first order reaction, rate constant can be calculated as follows:
$k = {{0.693} \over {{t_{1/2}}}}$ ..... (i)
Here, k is the rate constant and t$_{1/2}$ is the half life time.
As given, $t_{1/2}=100$ min
Substitute the respective value in equation (i), we get
$k = {{0.693} \over {100}}$
$\therefore~k=6.93\times10^{-3}$ min$^{-1}$
The rate constant for the reaction is $6.93\times10^{-3}$ min$^{-1}$.
(C) The time taken for completion of 75% reaction is 2$t_{1/2}$.
(D) For the given chemical reaction,
| At $t=0$ | 800 | 0 | 0 |
|---|---|---|---|
| After sometime | 800-2x | 3x | 2x |
The partial pressure at t = 0 and after some time is given below.
P$_{total}$ = 800 $-$ 2x + 3x + 2x
= 800 + 3x
Given, 800 $-$ 2x = 700
x = 50 mm
P$_{total}$ = 800 + 3 $\times$ 50 = 950 mm
The total pressure of the reaction mixture when $p_x=700$ mm.
Final Answer:
(A) First order reaction
(B) Rate constant = 6.93 $\times$ 10$^{–3} min$^–1}$
(C) Time taken for completion of 75% reaction is 2$t_{1/2}$
(D) Total pressure when $p_x = 700$ mm is 950 mm.
Shortcut Method:
Alternatively, the rate constant for the first order reaction can be calculated as follows:
Hence, for first order $k = {{2.303} \over t}{\log _{10}}{{{p_0}} \over {{p_1}}}$
At t = 100 min ${k_{100}} = {{2.303} \over {100}}{\log _{10}}{{800} \over {400}}$
$ = {{2.303} \over {100}}{\log _{10}}2 = {{2.303 \times 0.3010} \over {100}}$ min$^{-1}$
$ = 6.93 \times {10^{ - 3}}$ min$^{-1}$
| [Ao] | [Bo] | Ro (mol L-1 s-1) | |
|---|---|---|---|
| 1 | 0.1 | 0.1 | 0.05 |
| 2 | 0.2 | 0.1 | 0.10 |
| 3 | 0.1 | 0.2 | 0.05 |
(b) Find the rate constant
Explanation:
According to the given data, we see that when the concentration of A alone is doubled (0.1 $\to$ 0.2 mol L$-$1), the rate of the reaction gets doubled (0.05 $\to$ 0.1 mol L$-$1 s$-$1). Therefore, the order with respect to A is 1.
When the concentration of B alone is doubled (0.1 $\to$ 0.2 mol L$-$1), the rate of the reaction remains unaltered. Therefore, the order with respect to B is 0.
(1) The rate equation is, Rate = $k[A]{[B]^0} = k[A]$
(2) Rate $ = k[A]$ or, $k = {{rate} \over {[A]}}$ or, $k = {{0.05} \over {0.1}}$ $\therefore$ k = 0.5 s$-$1
Explanation:
For a first order reaction, rate = $k[A]$ . If A1 and A2 be the initial and final concentrations of the reactant respectively then,
${r_1} = k[{A_1}]$ or, $0.04 = k[{A_1}]$ .............. [1]
and ${r_2} = k[{A_2}]$ or, $0.03 = k[{A_2}]$ ....... [2]
$\therefore$ ${{[{A_1}]} \over {[{A_2}]}} = {{0.04} \over {0.03}} = {4 \over 3}$ (Dividing equation [1] by equation [2])
At 10 min, $10 = {{2.303} \over k}\log {{[{A_0}]} \over {[{A_1}]}}$, at 20 min, $20 = {{2.303} \over k}\log {{[{A_0}]} \over {[{A_2}]}}$
$\therefore$ $20 - 10 = 10 = {{2.303} \over k}\log {{[{A_1}]} \over {[{A_2}]}}$
or, $k = {{2.303} \over {10}}\log \left( {{4 \over 3}} \right)$ or, $k = 2.855 \times {10^{ - 2}}$ min$-$1
$\therefore$ The half-life for the first order reaction,
$ = {{0.693} \over k} = {{0.693} \over {2.855 \times {{10}^{ - 2}}}} = 24.27$ min
Explanation:
Given, initial moles of A = 10; initial moles of B = 12. Let, the number of moles of A = n, when polymerization is arrested. Moles of solute added = 0.525.
$\therefore$ Total number of moles = (n + 12 + 0.525) = (n + 12.525)
$\therefore$ Mole fraction of $A({x_A}) = {n \over {n + 12.525}}$ and
Mole fraction of $B({x_B}) = {{12} \over {n + 12.525}}$
The total vapour pressure of the solution = $P_A^0{X_A} + P_B^0{X_B}$
or, $400 = 300 \times {n \over {n + 12.525}} + 500 \times {{12} \over {n + 12.525}}$ or, n = 9.9
For a first order reaction, $k = {{2.303} \over t}\log {{{{[A]}_0}} \over {[A]}}$
or, $k = {{2.303} \over {100}}\log {a \over {a - x}} = {{2.303} \over {100}}\log {{10} \over {9.9}}$ [$\because$ a $-$ x = n = 9.9]
or, $k = 1.004 \times {10^{ - 4}}$ min$-$1
Explanation:
From Arrhenius equation, $\log k = \log A - {{{E_a}} \over {2.303RT}}$
$\therefore$ $\log {k_{500}} = \log A - {{{E_{{a_1}}}} \over {2.303R{T_1}}}$, $\log {k_{400}} = \log A - {{{E_{{a_2}}}} \over {2.303R{T_2}}}$
Given, ${k_{500}} = {k_{400}}$ $\therefore$ $\log {k_{500}} = \log {k_{400}}$
or, ${{{E_{{a_1}}}} \over {{T_1}}} = {{{E_{{a_2}}}} \over {{T_2}}}$ or, ${{{E_{{a_1}}}} \over {500}} = {{{E_{{a_2}}}} \over {400}}$ or, ${{{E_{{a_1}}}} \over {{E_{{a_2}}}}} = {{500} \over {400}} = {5 \over 4}$ ...... [1]
According to given data, ${E_{{a_1}}} - {E_{{a_2}}} = 20$
$\therefore$ Substituting in equation [1], we get ${{{E_{{a_1}}}} \over {{E_{{a_1}}} - 20}} = {5 \over 4}$
or, $4{E_{{a_1}}} - 5{E_{{a_1}}} - 100$ or, ${E_{{a_1}}} = 100$ kJ mol$-$1
$\therefore$ ${E_{{a_2}}} = 100 - 20 = 80$ kJ mol$-$1
Explanation:
The unit of rate constant = min$-$1 implies that the reaction is of first order.
For a first order reaction, $k = {{2.303} \over t}\log {a \over {a - x}}$
or, $k = {{2.303} \over t}\log {{{{[A]}_0}} \over {[A]}}$ or, $4.5 \times {10^{ - 3}} = {{2.303} \over {60}}\log {1 \over {[A]}}$
Hence, after 1 hour concentration of A, [ A ] = 0.764 (M)
$\therefore$ Reaction-rate after 1 hour = $k[A] = 4.5 \times {10^{ - 3}} \times 0.746$
= 3.438 $\times$ 10$-$3 mol L$-$1min$-$1
Explanation:
According to Arrhenius equation, $k = A{e^{ - {E_a}/RT}}$
$\therefore$ $\log k = \log A - {{{E_a}} \over {2.303RT}}$ or, $\log \left( {{{{k_2}} \over {{k_1}}}} \right) = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$
Given : $\log \left( {{{4.5 \times {{10}^7}} \over {1.5 \times {{10}^7}}}} \right) = {{{E_a}} \over {2.303 \times 8.314}}\left[ {{1 \over {323}} - {1 \over {373}}} \right]$
or, ${E_a} = 2.2 \times {10^4}$ J mol$-$1
From Arrhenius equation, $k = A{e^{ - {E_a}/RT}}$
or, $\log k = \log A - {{{E_a}} \over {2.303RT}}$
or, $\log (4.5 \times {10^7}) = \log A - {{2.2 \times {{10}^4}} \over {2.303 \times 8.314 \times 373}}$
or, $A = 5.42 \times {10^{10}}$ s$-$1.
Explanation:
Given, NH$_4^ + $ + H2O $\rightleftharpoons$ NH3 + H3O+
${K_a} = {{[N{H_3}][{H_3}{O^ + }]} \over {[NH_4^ + ][{H_2}O]}}$ (Ka = 5.6 $\times$ 10$-$10) ..... [1]
$NH_4^ + + \mathop O\limits^ - H$ $\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{{k_b}}^{{k_f}}} $ $N{H_3} + {H_2}O$ (kf = 3.4 $\times$ 1010 Lmol$-$1 s$-$1) ...... [2]
${K_{eq}} = {{{k_f}} \over {{k_b}}} = {{[N{H_3}][{H_3}O]} \over {[NH_4^ + ][O{H^ - }]}}$
$ = {{[N{H_3}][{H_3}{O^ + }]} \over {[NH_4^ + ][{H_2}O]}} \times {{{{[{H_2}O]}^2}} \over {[{H_3}{O^ + }][O{H^ - }]}} = {{{K_a}} \over {{K_w}}}$
$\therefore$ ${k_b} = {{{K_w}} \over {{K_a}}} \times {k_f} = {{1.0 \times {{10}^{ - 4}}} \over {5.6 \times {{10}^{ - 10}}}} \times 3.4 \times {10^{10}} = 6.07 \times {10^5}$
Explanation:
For the first order reaction half-life = ${{0.693} \over k}$
or, $360 = {{0.693} \over k}$ $\therefore$ ${k_{380^\circ C}} = {{0.693} \over {360}}$
We know, $\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$
$\therefore$ $\log {{{k_{450^\circ C}}} \over {{{0.693} \over {360}}}} = {{200 \times 1000} \over {2.303 \times 8.314}}\left[ {{1 \over {653}} - {1 \over {723}}} \right]$
or, ${k_{450^\circ C}} = 6.18 \times {10^{ - 2}}$ min$-$1
For 75% decomposition at 723K (450$^\circ$C),
${k_{450^\circ C}} = {{2.303} \over t}\log {a \over {a - x}}$
or, $6.81 \times {10^{ - 2}} = {{2.303} \over t}\log {{100} \over {(100 - 75)}}$ or, t = 20.358 min
Explanation:
From Arrhenius equation, $\log k = \log A - {{{E_a}} \over {2.303RT}}$
or, $\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$
$\therefore$ $\log {{{k_2}} \over {{k_1}}} = {{70} \over {2.303 \times 8.314}}\left[ {{1 \over {298}} - {1 \over {313}}} \right]$ or, ${{{k_2}} \over {{k_1}}} = 3.872$ ...... [1]
For a first order reaction, $k = {{2.303} \over t}\log {a \over {a - x}}$
Given x = 0.25 a,
$\therefore$ a $-$ x = (a $-$ 0.25 a) = 0.75 a at t = 20 min
$\therefore$ ${k_1} = {{2.303} \over {20}}\log {a \over {0.75\,a}} = 0.014386$ min$-$1 ...... [2]
Substituting for k1 in equation 1, we have
${k_2} = 3.872 \times 0.014386 = 0.05571$ min$-$1
For a first order reaction, ${k_2} = {{2.303} \over t}\log {a \over {a - x}}$
or, $0.05571 = {{2.303} \over {20}}\log {a \over {a - x}}$ [x = decrease in concentration of reactant]
or, $\log {a \over {a - x}} = {{0.05571 \times 20} \over {2.303}}$
or, $\log {a \over {a - x}} = 0.48381$
or, x = 0.6717 a
Hence, percentage decomposition $ = {{0.6717\,a} \over a} \times 100 = 67.17\% $