ChemistryQ151–156 of 194 questions
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151
ChemistryEasyMCQTS EAMCET · 2013
Which one of the following is an example of a disproportionation reaction?
A
$3Cl_{2\text{(g)}} + 6OH^-{_{\text{(aq)}}} \rightarrow ClO_3^-{_{\text{(aq)}}} + 5Cl^-{_{\text{(aq)}}} + 3H_2O_{\text{(l)}}$
B
$Ag^{2+}_{(aq)} + Ag_{(s)} \longrightarrow 2 Ag^+_{(aq)}$
C
$Zn_{(s)} + CuSO_{4(aq)} \longrightarrow Cu_{(s)} + ZnSO_{4(aq)}$
D
$2 KClO_{3(s)} \longrightarrow 2 KCl_{(s)} + 3 O_{2(g)}$
Solution
(A) disproportionation reaction is a special type of redox reaction in which the same element in a given oxidation state is simultaneously oxidized and reduced.
In the reaction $3Cl_{2\text{(g)}} + 6OH^-{_{\text{(aq)}}} \rightarrow ClO_3^-{_{\text{(aq)}}} + 5Cl^-{_{\text{(aq)}}} + 3H_2O_{\text{(l)}}$,the oxidation state of chlorine changes from $0$ in $Cl_2$ to $-1$ in $Cl^-$ (reduction) and to $+5$ in $ClO_3^-$ (oxidation).
Thus,it is a disproportionation reaction.
In the reaction $3Cl_{2\text{(g)}} + 6OH^-{_{\text{(aq)}}} \rightarrow ClO_3^-{_{\text{(aq)}}} + 5Cl^-{_{\text{(aq)}}} + 3H_2O_{\text{(l)}}$,the oxidation state of chlorine changes from $0$ in $Cl_2$ to $-1$ in $Cl^-$ (reduction) and to $+5$ in $ClO_3^-$ (oxidation).
Thus,it is a disproportionation reaction.
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152
ChemistryMediumMCQTS EAMCET · 2013
The degree of ionization of $0.10 \ M$ lactic acid is $4.0 \%$. The value of $K_a$ is:
A
$1.66 \times 10^{-5}$
B
$1.66 \times 10^{-4}$
C
$1.66 \times 10^{-3}$
D
$1.66 \times 10^{-2}$
Solution
(B) Given: Concentration $C = 0.10 \ M$,degree of ionization $\alpha = 4.0 \% = 0.04$.
For the dissociation of lactic acid $(CH_3CH(OH)COOH)$:
$CH_3CH(OH)COOH \rightleftharpoons CH_3CH(OH)COO^- + H^+$
At equilibrium,the concentrations are:
$[CH_3CH(OH)COOH] = C(1 - \alpha)$
$[CH_3CH(OH)COO^-] = C\alpha$
$[H^+] = C\alpha$
The dissociation constant $K_a$ is given by:
$K_a = \frac{[CH_3CH(OH)COO^-][H^+]}{[CH_3CH(OH)COOH]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha}$
Substituting the values:
$K_a = \frac{0.1 \times (0.04)^2}{1 - 0.04} = \frac{0.1 \times 0.0016}{0.96} = \frac{0.00016}{0.96} = \frac{1.6 \times 10^{-4}}{0.96} \approx 1.66 \times 10^{-4}$
For the dissociation of lactic acid $(CH_3CH(OH)COOH)$:
$CH_3CH(OH)COOH \rightleftharpoons CH_3CH(OH)COO^- + H^+$
At equilibrium,the concentrations are:
$[CH_3CH(OH)COOH] = C(1 - \alpha)$
$[CH_3CH(OH)COO^-] = C\alpha$
$[H^+] = C\alpha$
The dissociation constant $K_a$ is given by:
$K_a = \frac{[CH_3CH(OH)COO^-][H^+]}{[CH_3CH(OH)COOH]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha}$
Substituting the values:
$K_a = \frac{0.1 \times (0.04)^2}{1 - 0.04} = \frac{0.1 \times 0.0016}{0.96} = \frac{0.00016}{0.96} = \frac{1.6 \times 10^{-4}}{0.96} \approx 1.66 \times 10^{-4}$
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153
ChemistryEasyMCQTS EAMCET · 2013
At $T$ $K$,the ratio of kinetic energies of $4$ $g$ of $H_{2(g)}$ and $8$ $g$ of $O_{2(g)}$ is
A
$1: 4$
B
$4: 1$
C
$2: 1$
D
$8: 1$
Solution
(D) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = n \times \frac{3}{2} RT$.
For $4$ $g$ of $H_2$,the number of moles $n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2$ $mol$.
So,$KE_{H_2} = 2 \times \frac{3}{2} RT = 3RT$.
For $8$ $g$ of $O_2$,the number of moles $n_{O_2} = \frac{8 \ g}{32 \ g/mol} = 0.25$ $mol$ or $\frac{1}{4}$ $mol$.
So,$KE_{O_2} = \frac{1}{4} \times \frac{3}{2} RT = \frac{3}{8} RT$.
The ratio $KE_{H_2} : KE_{O_2} = 3RT : \frac{3}{8} RT = 1 : \frac{1}{8} = 8 : 1$.
For $4$ $g$ of $H_2$,the number of moles $n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2$ $mol$.
So,$KE_{H_2} = 2 \times \frac{3}{2} RT = 3RT$.
For $8$ $g$ of $O_2$,the number of moles $n_{O_2} = \frac{8 \ g}{32 \ g/mol} = 0.25$ $mol$ or $\frac{1}{4}$ $mol$.
So,$KE_{O_2} = \frac{1}{4} \times \frac{3}{2} RT = \frac{3}{8} RT$.
The ratio $KE_{H_2} : KE_{O_2} = 3RT : \frac{3}{8} RT = 1 : \frac{1}{8} = 8 : 1$.
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154
ChemistryMediumMCQTS EAMCET · 2013
The basis of the quantum mechanical model of an atom is:
A
angular momentum of electron
B
quantum numbers
C
dual nature of electron
D
black body radiation
Solution
(C) The quantum mechanical model of an atom is based on the dual nature of matter (wave-particle duality) proposed by $de \text{ } Broglie$.
This model considers the electron as both a particle and a wave,which is mathematically described by the $Schrodinger$ wave equation.
This model considers the electron as both a particle and a wave,which is mathematically described by the $Schrodinger$ wave equation.
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155
ChemistryMCQTS EAMCET · 2013
The number of radial nodes of $3s$ and $2p$ orbitals respectively are
A
$0, 2$
B
$2, 0$
C
$1, 2$
D
$2, 1$
Solution
(B) The number of radial nodes in an orbital is given by the formula: $n - l - 1$.
For $3s$ orbital: $n = 3$ and $l = 0$. Therefore,the number of radial nodes $= 3 - 0 - 1 = 2$.
For $2p$ orbital: $n = 2$ and $l = 1$. Therefore,the number of radial nodes $= 2 - 1 - 1 = 0$.
Thus,the number of radial nodes for $3s$ and $2p$ orbitals are $2$ and $0$ respectively.
For $3s$ orbital: $n = 3$ and $l = 0$. Therefore,the number of radial nodes $= 3 - 0 - 1 = 2$.
For $2p$ orbital: $n = 2$ and $l = 1$. Therefore,the number of radial nodes $= 2 - 1 - 1 = 0$.
Thus,the number of radial nodes for $3s$ and $2p$ orbitals are $2$ and $0$ respectively.
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156
ChemistryEasyMCQTS EAMCET · 2013
For which one of the following reactions,the entropy change is positive?
A
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$
B
$Na^{+}_{(g)} + Cl^{-}_{(g)} \longrightarrow NaCl_{(s)}$
C
$NaCl_{(l)} \longrightarrow NaCl_{(s)}$
D
$H_2O_{(l)} \longrightarrow H_2O_{(g)}$
Solution
(D) Entropy $(S)$ is a measure of the randomness or disorder of a system. $A$ positive entropy change $(\Delta S > 0)$ occurs when the system becomes more disordered,such as during a phase transition from a more ordered state (liquid) to a less ordered state (gas).
In the reaction $H_2O_{(l)} \longrightarrow H_2O_{(g)}$,liquid water is converted into water vapor. Since gas molecules have much higher randomness compared to liquid molecules,the entropy of the system increases,making $\Delta S$ positive.
In other options,the reactions involve a decrease in the number of moles of gas or a transition from a disordered state to a more ordered state (like liquid to solid or gas to solid),which results in a negative entropy change.
In the reaction $H_2O_{(l)} \longrightarrow H_2O_{(g)}$,liquid water is converted into water vapor. Since gas molecules have much higher randomness compared to liquid molecules,the entropy of the system increases,making $\Delta S$ positive.
In other options,the reactions involve a decrease in the number of moles of gas or a transition from a disordered state to a more ordered state (like liquid to solid or gas to solid),which results in a negative entropy change.
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