ChemistryQ1–6 of 6 questions
Page 1 of 1 · English
1
ChemistryEasyMCQGUJCET · 2008
Potassium dichromate $(K_2Cr_2O_7)$ is used . . . . . . .
A
as a reducing agent
B
in electroplating
C
as an insecticide
D
as an oxidizing agent to oxidize ferrous ions $(Fe^{2+})$ into ferric ions $(Fe^{3+})$ in acidic media.
Solution
(D) Potassium dichromate $(K_2Cr_2O_7)$ is a strong oxidizing agent in acidic medium.
It reacts with ferrous ions $(Fe^{2+})$ to oxidize them into ferric ions $(Fe^{3+})$ according to the following reaction:
$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
Therefore,the correct statement is that it acts as an oxidizing agent in acidic media.
It reacts with ferrous ions $(Fe^{2+})$ to oxidize them into ferric ions $(Fe^{3+})$ according to the following reaction:
$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
Therefore,the correct statement is that it acts as an oxidizing agent in acidic media.
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2
ChemistryEasyMCQGUJCET · 2008
When a graph of concentration $\rightarrow$ Time is plotted for a zero order reaction,the value of the slope is . . . . . . .
A
$-2.303 \times k$
B
$-\frac{k}{2.303}$
C
$-\frac{E_a}{2.303R}$
D
$-k$
Solution
(D) For a zero order reaction,the integrated rate equation is given by: $[A] = [A]_0 - kt$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = [A]$,$x = t$,$m = -k$,and $c = [A]_0$.
Thus,the slope of the graph of concentration $[A]$ versus time $t$ is equal to $-k$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = [A]$,$x = t$,$m = -k$,and $c = [A]_0$.
Thus,the slope of the graph of concentration $[A]$ versus time $t$ is equal to $-k$.
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3
ChemistryEasyMCQGUJCET · 2008
Which equation is true to calculate the energy of activation,if the rate of reaction is doubled by increasing temperature from $T_1 \ K$ to $T_2 \ K$?
A
$\log_{10} \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
B
$\log_{10} 2 = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
C
$\log_{10} \frac{k_1}{k_2} = \frac{E_a}{2.303R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$
D
$\log_{10} \frac{1}{2} = \frac{E_a}{2.303} \left[ \frac{1}{T_2} - \frac{1}{T_1} \right]$
Solution
(B) The Arrhenius equation for two different temperatures is given by: $\log_{10} \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$.
Given that the rate of reaction is doubled,the rate constant $k_2 = 2k_1$,which implies $\frac{k_2}{k_1} = 2$.
Substituting this into the Arrhenius equation,we get: $\log_{10} 2 = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$.
Therefore,option $B$ is the correct equation.
Given that the rate of reaction is doubled,the rate constant $k_2 = 2k_1$,which implies $\frac{k_2}{k_1} = 2$.
Substituting this into the Arrhenius equation,we get: $\log_{10} 2 = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$.
Therefore,option $B$ is the correct equation.
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4
ChemistryEasyMCQGUJCET · 2008
The order of a reaction for an esterification process is . . . . . . .
A
First
B
Zero
C
Pseudo First order
D
Second order
Solution
(C) The esterification reaction involves the reaction between an ester and water in the presence of an acid catalyst: $RCOOR' + H_2O \xrightarrow{H^+} RCOOH + R'OH$.
Since water is present in large excess,its concentration remains practically constant during the reaction.
Therefore,the rate of the reaction depends only on the concentration of the ester.
Such reactions,which are theoretically second-order but behave as first-order,are known as pseudo-first-order reactions.
Since water is present in large excess,its concentration remains practically constant during the reaction.
Therefore,the rate of the reaction depends only on the concentration of the ester.
Such reactions,which are theoretically second-order but behave as first-order,are known as pseudo-first-order reactions.
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5
ChemistryEasyMCQGUJCET · 2008
Which ligand is useful for the removal of the toxic effect of lead metal in the body during chelate therapy treatment?
A
$EDTA$
B
$C_2O_4^{2-}$
C
$AsO_4^{3-}$
D
$EDTA^{4-}$ structure
Solution
(A) $EDTA$ (Ethylenediaminetetraacetate) is a hexadentate ligand that forms stable,water-soluble complexes with metal ions like lead $(Pb^{2+})$.
In chelate therapy,$EDTA$ is administered to bind with lead ions in the body,forming a stable complex that can be easily excreted through urine,thereby reducing the toxic effects of lead.
In chelate therapy,$EDTA$ is administered to bind with lead ions in the body,forming a stable complex that can be easily excreted through urine,thereby reducing the toxic effects of lead.
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6
ChemistryEasyMCQGUJCET · 2008
On which factors does the stability of an oxidation state in Lanthanoid elements depend?
A
Internal energy
B
Enthalpy
C
Electronic configuration
D
Combined effect of hydration energy and ionization energy
Solution
(D) The stability of a particular oxidation state in Lanthanoid elements is determined by the combined effect of the energy required to remove electrons (ionization energy) and the energy released when the resulting ions are hydrated (hydration energy).$ \Delta H_{total} = \Delta H_{ionization} + \Delta H_{hydration} $. Therefore,the correct option is $D$.
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