285424
The rate of a gaseous reaction is given by the expression, \(k[A][B]^2\). If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is
1 \(\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 8
4 16
Explanation:
(c) \(R=k[A]^1[B]^2\)
Volume is reduced to half means pressure is doubled.
\(R^{\prime}=k[2 A][2 B]^2\)
\(R^{\prime}=k[A][B]^2 \times 8\)
\(R^{\prime}=8 R\)
\(\therefore \quad\) Rate of reaction becomes 8 times the initial rate.
Karnataka CET 2021
CHEMISTRY(KCET)
285425
Higher order \((>3)\) reactions are rare due to
1 shifting of equilibrium towards reactants due to elastic collisions
2 loss of active species on collision
3 low probability of simultaneous collision of all reacting species
4 increase in entropy as more molecules are involved.
Explanation:
(c) Higher order reactions are rare because there is low probability of simultaneous collisions of all reacting species, so the number of effective collisions becomes very less (negligible).
Karnataka CET 2021
CHEMISTRY(KCET)
285426
The time required for \(60 \%\) completion of a first order reaction is 50 min . The time required for \(93.6 \%\) completion of the same reaction will be
1 83.8 min
2 50 min
3 150 min
4 100 min .
Explanation:
(c) For first order reaction,
\(k=\frac{2.303}{t} \log \frac{a}{a-x}\)
where, \(a=\) initial concentration of the reactant, \(a-x\) \(=\) concentration of the reactant after time, \(t\). For \(60 \%\) completion of reaction,
\(k=\frac{2.303}{50} \log \frac{100}{40}\)
\(k=\frac{2.303}{50} \times 0.397=0.0183\)
For \(93.6 \%\) completion of reaction,
\(k=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(0.0183=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(t=\frac{2.303}{0.0183} \times 1.193\)
\(t=150 \mathrm{~min}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285427
For an elementary reaction.
\(2 A+3 B \rightarrow 4 C+D\) the rate of appearance of \(C\) at time ' \(t\) ' is \(2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). Rate of disappearance of \(B\) at ' \(t\) ', \(t\) will be
285424
The rate of a gaseous reaction is given by the expression, \(k[A][B]^2\). If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is
1 \(\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 8
4 16
Explanation:
(c) \(R=k[A]^1[B]^2\)
Volume is reduced to half means pressure is doubled.
\(R^{\prime}=k[2 A][2 B]^2\)
\(R^{\prime}=k[A][B]^2 \times 8\)
\(R^{\prime}=8 R\)
\(\therefore \quad\) Rate of reaction becomes 8 times the initial rate.
Karnataka CET 2021
CHEMISTRY(KCET)
285425
Higher order \((>3)\) reactions are rare due to
1 shifting of equilibrium towards reactants due to elastic collisions
2 loss of active species on collision
3 low probability of simultaneous collision of all reacting species
4 increase in entropy as more molecules are involved.
Explanation:
(c) Higher order reactions are rare because there is low probability of simultaneous collisions of all reacting species, so the number of effective collisions becomes very less (negligible).
Karnataka CET 2021
CHEMISTRY(KCET)
285426
The time required for \(60 \%\) completion of a first order reaction is 50 min . The time required for \(93.6 \%\) completion of the same reaction will be
1 83.8 min
2 50 min
3 150 min
4 100 min .
Explanation:
(c) For first order reaction,
\(k=\frac{2.303}{t} \log \frac{a}{a-x}\)
where, \(a=\) initial concentration of the reactant, \(a-x\) \(=\) concentration of the reactant after time, \(t\). For \(60 \%\) completion of reaction,
\(k=\frac{2.303}{50} \log \frac{100}{40}\)
\(k=\frac{2.303}{50} \times 0.397=0.0183\)
For \(93.6 \%\) completion of reaction,
\(k=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(0.0183=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(t=\frac{2.303}{0.0183} \times 1.193\)
\(t=150 \mathrm{~min}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285427
For an elementary reaction.
\(2 A+3 B \rightarrow 4 C+D\) the rate of appearance of \(C\) at time ' \(t\) ' is \(2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). Rate of disappearance of \(B\) at ' \(t\) ', \(t\) will be
285424
The rate of a gaseous reaction is given by the expression, \(k[A][B]^2\). If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is
1 \(\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 8
4 16
Explanation:
(c) \(R=k[A]^1[B]^2\)
Volume is reduced to half means pressure is doubled.
\(R^{\prime}=k[2 A][2 B]^2\)
\(R^{\prime}=k[A][B]^2 \times 8\)
\(R^{\prime}=8 R\)
\(\therefore \quad\) Rate of reaction becomes 8 times the initial rate.
Karnataka CET 2021
CHEMISTRY(KCET)
285425
Higher order \((>3)\) reactions are rare due to
1 shifting of equilibrium towards reactants due to elastic collisions
2 loss of active species on collision
3 low probability of simultaneous collision of all reacting species
4 increase in entropy as more molecules are involved.
Explanation:
(c) Higher order reactions are rare because there is low probability of simultaneous collisions of all reacting species, so the number of effective collisions becomes very less (negligible).
Karnataka CET 2021
CHEMISTRY(KCET)
285426
The time required for \(60 \%\) completion of a first order reaction is 50 min . The time required for \(93.6 \%\) completion of the same reaction will be
1 83.8 min
2 50 min
3 150 min
4 100 min .
Explanation:
(c) For first order reaction,
\(k=\frac{2.303}{t} \log \frac{a}{a-x}\)
where, \(a=\) initial concentration of the reactant, \(a-x\) \(=\) concentration of the reactant after time, \(t\). For \(60 \%\) completion of reaction,
\(k=\frac{2.303}{50} \log \frac{100}{40}\)
\(k=\frac{2.303}{50} \times 0.397=0.0183\)
For \(93.6 \%\) completion of reaction,
\(k=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(0.0183=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(t=\frac{2.303}{0.0183} \times 1.193\)
\(t=150 \mathrm{~min}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285427
For an elementary reaction.
\(2 A+3 B \rightarrow 4 C+D\) the rate of appearance of \(C\) at time ' \(t\) ' is \(2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). Rate of disappearance of \(B\) at ' \(t\) ', \(t\) will be
285424
The rate of a gaseous reaction is given by the expression, \(k[A][B]^2\). If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is
1 \(\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 8
4 16
Explanation:
(c) \(R=k[A]^1[B]^2\)
Volume is reduced to half means pressure is doubled.
\(R^{\prime}=k[2 A][2 B]^2\)
\(R^{\prime}=k[A][B]^2 \times 8\)
\(R^{\prime}=8 R\)
\(\therefore \quad\) Rate of reaction becomes 8 times the initial rate.
Karnataka CET 2021
CHEMISTRY(KCET)
285425
Higher order \((>3)\) reactions are rare due to
1 shifting of equilibrium towards reactants due to elastic collisions
2 loss of active species on collision
3 low probability of simultaneous collision of all reacting species
4 increase in entropy as more molecules are involved.
Explanation:
(c) Higher order reactions are rare because there is low probability of simultaneous collisions of all reacting species, so the number of effective collisions becomes very less (negligible).
Karnataka CET 2021
CHEMISTRY(KCET)
285426
The time required for \(60 \%\) completion of a first order reaction is 50 min . The time required for \(93.6 \%\) completion of the same reaction will be
1 83.8 min
2 50 min
3 150 min
4 100 min .
Explanation:
(c) For first order reaction,
\(k=\frac{2.303}{t} \log \frac{a}{a-x}\)
where, \(a=\) initial concentration of the reactant, \(a-x\) \(=\) concentration of the reactant after time, \(t\). For \(60 \%\) completion of reaction,
\(k=\frac{2.303}{50} \log \frac{100}{40}\)
\(k=\frac{2.303}{50} \times 0.397=0.0183\)
For \(93.6 \%\) completion of reaction,
\(k=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(0.0183=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(t=\frac{2.303}{0.0183} \times 1.193\)
\(t=150 \mathrm{~min}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285427
For an elementary reaction.
\(2 A+3 B \rightarrow 4 C+D\) the rate of appearance of \(C\) at time ' \(t\) ' is \(2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). Rate of disappearance of \(B\) at ' \(t\) ', \(t\) will be
285424
The rate of a gaseous reaction is given by the expression, \(k[A][B]^2\). If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is
1 \(\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 8
4 16
Explanation:
(c) \(R=k[A]^1[B]^2\)
Volume is reduced to half means pressure is doubled.
\(R^{\prime}=k[2 A][2 B]^2\)
\(R^{\prime}=k[A][B]^2 \times 8\)
\(R^{\prime}=8 R\)
\(\therefore \quad\) Rate of reaction becomes 8 times the initial rate.
Karnataka CET 2021
CHEMISTRY(KCET)
285425
Higher order \((>3)\) reactions are rare due to
1 shifting of equilibrium towards reactants due to elastic collisions
2 loss of active species on collision
3 low probability of simultaneous collision of all reacting species
4 increase in entropy as more molecules are involved.
Explanation:
(c) Higher order reactions are rare because there is low probability of simultaneous collisions of all reacting species, so the number of effective collisions becomes very less (negligible).
Karnataka CET 2021
CHEMISTRY(KCET)
285426
The time required for \(60 \%\) completion of a first order reaction is 50 min . The time required for \(93.6 \%\) completion of the same reaction will be
1 83.8 min
2 50 min
3 150 min
4 100 min .
Explanation:
(c) For first order reaction,
\(k=\frac{2.303}{t} \log \frac{a}{a-x}\)
where, \(a=\) initial concentration of the reactant, \(a-x\) \(=\) concentration of the reactant after time, \(t\). For \(60 \%\) completion of reaction,
\(k=\frac{2.303}{50} \log \frac{100}{40}\)
\(k=\frac{2.303}{50} \times 0.397=0.0183\)
For \(93.6 \%\) completion of reaction,
\(k=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(0.0183=\frac{2.303}{t} \log \frac{100}{6.4}\)
\(t=\frac{2.303}{0.0183} \times 1.193\)
\(t=150 \mathrm{~min}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285427
For an elementary reaction.
\(2 A+3 B \rightarrow 4 C+D\) the rate of appearance of \(C\) at time ' \(t\) ' is \(2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). Rate of disappearance of \(B\) at ' \(t\) ', \(t\) will be
