NEET Test Series from KOTA - 10 Papers In MS WORD
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CHEMISTRY(KCET)
285428
The rate constant of a reaction is given by \(k=P Z e^{-E_\alpha / R T}\) under standard notation. In order to speed up the reaction, which of the following factors has to be decreased?
1 Both \(Z\) and \(T\)
2 \(E_a\)
3 \(T\)
4 \(Z\)
Explanation:
(b) \(k=P Z e^{-E_a / R T}\)
\(\log k=\log (P Z)-\frac{E_a}{R T}\)
For \(k\) to increase, \(-\frac{E_a}{R T}\) should be small and that is possible only when \(E_a\) is less or \(T\) is more. So, for the reaction to speed up rapidly \(E_a\) should decrease.
Karnataka CET 2020
CHEMISTRY(KCET)
285431
1 L of \(2 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}\) is mixed with 1 L of 3 M \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\) to form an ester. The rate of the reaction with respect to the initial rate when each solution is diluted with an equal volume of water will be
1 0.25 times
2 2 times
3 0.5 times
4 4 times.
Explanation:
(a) \(r=k\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]\)
\(r_i=k[2 \mathrm{M}][3 \mathrm{M}]\)
\(r_f=k\left[\frac{2 \mathrm{M}}{2}\right]\left[\frac{3 \mathrm{M}}{2}\right]\)
On dividing (ii) by (i), we get
\(\frac{r_f}{r_i}=\frac{k[\mathrm{M}]\left[\frac{3}{2} \mathrm{M}\right]}{k[2 \mathrm{M}][3 \mathrm{M}]}=0.25\)
\(\therefore\) The rate of reaction becomes 0.25 times of initial rate.
Karnataka CET 2019
CHEMISTRY(KCET)
285432
For the reaction, \(2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3\), the rate of disappearance of \(\mathrm{O}_2\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). The rate of appearance of \(\mathrm{SO}_3\) is
285433
The temperature coefficient of a reaction is 2 . When the temperature is increased from \(30^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the rate of reaction is increased by
1 150 times
2 410 times
3 72 times
4 64 times.
Explanation:
(d) Temperature coefficient 2 means rate of reaction doubles at every \(10^{\circ} \mathrm{C}\) rise in temperature. Thus, \(\frac{k_{90^{\circ} \mathrm{C}}}{k_{30^{\circ} \mathrm{C}}}=2^6=64\)
285428
The rate constant of a reaction is given by \(k=P Z e^{-E_\alpha / R T}\) under standard notation. In order to speed up the reaction, which of the following factors has to be decreased?
1 Both \(Z\) and \(T\)
2 \(E_a\)
3 \(T\)
4 \(Z\)
Explanation:
(b) \(k=P Z e^{-E_a / R T}\)
\(\log k=\log (P Z)-\frac{E_a}{R T}\)
For \(k\) to increase, \(-\frac{E_a}{R T}\) should be small and that is possible only when \(E_a\) is less or \(T\) is more. So, for the reaction to speed up rapidly \(E_a\) should decrease.
Karnataka CET 2020
CHEMISTRY(KCET)
285431
1 L of \(2 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}\) is mixed with 1 L of 3 M \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\) to form an ester. The rate of the reaction with respect to the initial rate when each solution is diluted with an equal volume of water will be
1 0.25 times
2 2 times
3 0.5 times
4 4 times.
Explanation:
(a) \(r=k\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]\)
\(r_i=k[2 \mathrm{M}][3 \mathrm{M}]\)
\(r_f=k\left[\frac{2 \mathrm{M}}{2}\right]\left[\frac{3 \mathrm{M}}{2}\right]\)
On dividing (ii) by (i), we get
\(\frac{r_f}{r_i}=\frac{k[\mathrm{M}]\left[\frac{3}{2} \mathrm{M}\right]}{k[2 \mathrm{M}][3 \mathrm{M}]}=0.25\)
\(\therefore\) The rate of reaction becomes 0.25 times of initial rate.
Karnataka CET 2019
CHEMISTRY(KCET)
285432
For the reaction, \(2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3\), the rate of disappearance of \(\mathrm{O}_2\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). The rate of appearance of \(\mathrm{SO}_3\) is
285433
The temperature coefficient of a reaction is 2 . When the temperature is increased from \(30^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the rate of reaction is increased by
1 150 times
2 410 times
3 72 times
4 64 times.
Explanation:
(d) Temperature coefficient 2 means rate of reaction doubles at every \(10^{\circ} \mathrm{C}\) rise in temperature. Thus, \(\frac{k_{90^{\circ} \mathrm{C}}}{k_{30^{\circ} \mathrm{C}}}=2^6=64\)
285428
The rate constant of a reaction is given by \(k=P Z e^{-E_\alpha / R T}\) under standard notation. In order to speed up the reaction, which of the following factors has to be decreased?
1 Both \(Z\) and \(T\)
2 \(E_a\)
3 \(T\)
4 \(Z\)
Explanation:
(b) \(k=P Z e^{-E_a / R T}\)
\(\log k=\log (P Z)-\frac{E_a}{R T}\)
For \(k\) to increase, \(-\frac{E_a}{R T}\) should be small and that is possible only when \(E_a\) is less or \(T\) is more. So, for the reaction to speed up rapidly \(E_a\) should decrease.
Karnataka CET 2020
CHEMISTRY(KCET)
285431
1 L of \(2 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}\) is mixed with 1 L of 3 M \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\) to form an ester. The rate of the reaction with respect to the initial rate when each solution is diluted with an equal volume of water will be
1 0.25 times
2 2 times
3 0.5 times
4 4 times.
Explanation:
(a) \(r=k\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]\)
\(r_i=k[2 \mathrm{M}][3 \mathrm{M}]\)
\(r_f=k\left[\frac{2 \mathrm{M}}{2}\right]\left[\frac{3 \mathrm{M}}{2}\right]\)
On dividing (ii) by (i), we get
\(\frac{r_f}{r_i}=\frac{k[\mathrm{M}]\left[\frac{3}{2} \mathrm{M}\right]}{k[2 \mathrm{M}][3 \mathrm{M}]}=0.25\)
\(\therefore\) The rate of reaction becomes 0.25 times of initial rate.
Karnataka CET 2019
CHEMISTRY(KCET)
285432
For the reaction, \(2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3\), the rate of disappearance of \(\mathrm{O}_2\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). The rate of appearance of \(\mathrm{SO}_3\) is
285433
The temperature coefficient of a reaction is 2 . When the temperature is increased from \(30^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the rate of reaction is increased by
1 150 times
2 410 times
3 72 times
4 64 times.
Explanation:
(d) Temperature coefficient 2 means rate of reaction doubles at every \(10^{\circ} \mathrm{C}\) rise in temperature. Thus, \(\frac{k_{90^{\circ} \mathrm{C}}}{k_{30^{\circ} \mathrm{C}}}=2^6=64\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHEMISTRY(KCET)
285428
The rate constant of a reaction is given by \(k=P Z e^{-E_\alpha / R T}\) under standard notation. In order to speed up the reaction, which of the following factors has to be decreased?
1 Both \(Z\) and \(T\)
2 \(E_a\)
3 \(T\)
4 \(Z\)
Explanation:
(b) \(k=P Z e^{-E_a / R T}\)
\(\log k=\log (P Z)-\frac{E_a}{R T}\)
For \(k\) to increase, \(-\frac{E_a}{R T}\) should be small and that is possible only when \(E_a\) is less or \(T\) is more. So, for the reaction to speed up rapidly \(E_a\) should decrease.
Karnataka CET 2020
CHEMISTRY(KCET)
285431
1 L of \(2 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}\) is mixed with 1 L of 3 M \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\) to form an ester. The rate of the reaction with respect to the initial rate when each solution is diluted with an equal volume of water will be
1 0.25 times
2 2 times
3 0.5 times
4 4 times.
Explanation:
(a) \(r=k\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]\)
\(r_i=k[2 \mathrm{M}][3 \mathrm{M}]\)
\(r_f=k\left[\frac{2 \mathrm{M}}{2}\right]\left[\frac{3 \mathrm{M}}{2}\right]\)
On dividing (ii) by (i), we get
\(\frac{r_f}{r_i}=\frac{k[\mathrm{M}]\left[\frac{3}{2} \mathrm{M}\right]}{k[2 \mathrm{M}][3 \mathrm{M}]}=0.25\)
\(\therefore\) The rate of reaction becomes 0.25 times of initial rate.
Karnataka CET 2019
CHEMISTRY(KCET)
285432
For the reaction, \(2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3\), the rate of disappearance of \(\mathrm{O}_2\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). The rate of appearance of \(\mathrm{SO}_3\) is
285433
The temperature coefficient of a reaction is 2 . When the temperature is increased from \(30^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the rate of reaction is increased by
1 150 times
2 410 times
3 72 times
4 64 times.
Explanation:
(d) Temperature coefficient 2 means rate of reaction doubles at every \(10^{\circ} \mathrm{C}\) rise in temperature. Thus, \(\frac{k_{90^{\circ} \mathrm{C}}}{k_{30^{\circ} \mathrm{C}}}=2^6=64\)