285438
The half-life period of a first order reaction is 60 minutes. What percentage will be left over after 240 minutes?
1 \(6.25 \%\)
2 \(4.25 \%\)
3 \(5 \%\)
4 \(6 \%\)
Explanation:
(a) \(t_{1 / 2}=\frac{0.693}{k} \Rightarrow \frac{0.693}{t_{1 / 2}}=k \Rightarrow \frac{0.693}{60}=k\)
Chemical Kinetics
c. 71
\(k=0.01155 \mathrm{~min}^{-1}\)
\(k=\frac{2.303}{t} \log \left(\frac{a}{a-x}\right)\)
Let the initial amount (a) be 100 .
\(0.01155 \mathrm{~min}^{-1}=\frac{2.303}{240 \mathrm{~min}} \log \left(\frac{100}{a-x}\right)\).
\(\frac{0.01155 \mathrm{~min}^{-1} \times 240 \mathrm{~min}}{2.303}=\log \left(\frac{100}{a-x}\right)\)
\(1.204=\log 100-\log (a-x)\)
\(1.204=2-\log (a-x)\)
\(\log (a-x)=2-1.204\)
\(\log (a-x)=0.796\)
or \((a-x)=6.25 \%\)
Karnataka CET 2016
CHEMISTRY(KCET)
285439
For a chemical reaction, \(m A \rightarrow x B\), the rate law is \(r=k[A]^2\).
If the concentration of \(A\) is doubled, the reaction rate will be
1 doubled
2 quadrupled
3 increased by 8 times
4 unchanged.
Explanation:
(b) \(r=k[A]^2\)
When concentration of \(A\) is doubled,
\(r=k[2 A]^2\)
\(r=4 k[A]^2\)
Karnataka CET 2016
CHEMISTRY(KCET)
285440
\(3 A \rightarrow 2 B\), rate of reaction, \(+\frac{d[B]}{d t}\) is equal to
285438
The half-life period of a first order reaction is 60 minutes. What percentage will be left over after 240 minutes?
1 \(6.25 \%\)
2 \(4.25 \%\)
3 \(5 \%\)
4 \(6 \%\)
Explanation:
(a) \(t_{1 / 2}=\frac{0.693}{k} \Rightarrow \frac{0.693}{t_{1 / 2}}=k \Rightarrow \frac{0.693}{60}=k\)
Chemical Kinetics
c. 71
\(k=0.01155 \mathrm{~min}^{-1}\)
\(k=\frac{2.303}{t} \log \left(\frac{a}{a-x}\right)\)
Let the initial amount (a) be 100 .
\(0.01155 \mathrm{~min}^{-1}=\frac{2.303}{240 \mathrm{~min}} \log \left(\frac{100}{a-x}\right)\).
\(\frac{0.01155 \mathrm{~min}^{-1} \times 240 \mathrm{~min}}{2.303}=\log \left(\frac{100}{a-x}\right)\)
\(1.204=\log 100-\log (a-x)\)
\(1.204=2-\log (a-x)\)
\(\log (a-x)=2-1.204\)
\(\log (a-x)=0.796\)
or \((a-x)=6.25 \%\)
Karnataka CET 2016
CHEMISTRY(KCET)
285439
For a chemical reaction, \(m A \rightarrow x B\), the rate law is \(r=k[A]^2\).
If the concentration of \(A\) is doubled, the reaction rate will be
1 doubled
2 quadrupled
3 increased by 8 times
4 unchanged.
Explanation:
(b) \(r=k[A]^2\)
When concentration of \(A\) is doubled,
\(r=k[2 A]^2\)
\(r=4 k[A]^2\)
Karnataka CET 2016
CHEMISTRY(KCET)
285440
\(3 A \rightarrow 2 B\), rate of reaction, \(+\frac{d[B]}{d t}\) is equal to
285438
The half-life period of a first order reaction is 60 minutes. What percentage will be left over after 240 minutes?
1 \(6.25 \%\)
2 \(4.25 \%\)
3 \(5 \%\)
4 \(6 \%\)
Explanation:
(a) \(t_{1 / 2}=\frac{0.693}{k} \Rightarrow \frac{0.693}{t_{1 / 2}}=k \Rightarrow \frac{0.693}{60}=k\)
Chemical Kinetics
c. 71
\(k=0.01155 \mathrm{~min}^{-1}\)
\(k=\frac{2.303}{t} \log \left(\frac{a}{a-x}\right)\)
Let the initial amount (a) be 100 .
\(0.01155 \mathrm{~min}^{-1}=\frac{2.303}{240 \mathrm{~min}} \log \left(\frac{100}{a-x}\right)\).
\(\frac{0.01155 \mathrm{~min}^{-1} \times 240 \mathrm{~min}}{2.303}=\log \left(\frac{100}{a-x}\right)\)
\(1.204=\log 100-\log (a-x)\)
\(1.204=2-\log (a-x)\)
\(\log (a-x)=2-1.204\)
\(\log (a-x)=0.796\)
or \((a-x)=6.25 \%\)
Karnataka CET 2016
CHEMISTRY(KCET)
285439
For a chemical reaction, \(m A \rightarrow x B\), the rate law is \(r=k[A]^2\).
If the concentration of \(A\) is doubled, the reaction rate will be
1 doubled
2 quadrupled
3 increased by 8 times
4 unchanged.
Explanation:
(b) \(r=k[A]^2\)
When concentration of \(A\) is doubled,
\(r=k[2 A]^2\)
\(r=4 k[A]^2\)
Karnataka CET 2016
CHEMISTRY(KCET)
285440
\(3 A \rightarrow 2 B\), rate of reaction, \(+\frac{d[B]}{d t}\) is equal to
285438
The half-life period of a first order reaction is 60 minutes. What percentage will be left over after 240 minutes?
1 \(6.25 \%\)
2 \(4.25 \%\)
3 \(5 \%\)
4 \(6 \%\)
Explanation:
(a) \(t_{1 / 2}=\frac{0.693}{k} \Rightarrow \frac{0.693}{t_{1 / 2}}=k \Rightarrow \frac{0.693}{60}=k\)
Chemical Kinetics
c. 71
\(k=0.01155 \mathrm{~min}^{-1}\)
\(k=\frac{2.303}{t} \log \left(\frac{a}{a-x}\right)\)
Let the initial amount (a) be 100 .
\(0.01155 \mathrm{~min}^{-1}=\frac{2.303}{240 \mathrm{~min}} \log \left(\frac{100}{a-x}\right)\).
\(\frac{0.01155 \mathrm{~min}^{-1} \times 240 \mathrm{~min}}{2.303}=\log \left(\frac{100}{a-x}\right)\)
\(1.204=\log 100-\log (a-x)\)
\(1.204=2-\log (a-x)\)
\(\log (a-x)=2-1.204\)
\(\log (a-x)=0.796\)
or \((a-x)=6.25 \%\)
Karnataka CET 2016
CHEMISTRY(KCET)
285439
For a chemical reaction, \(m A \rightarrow x B\), the rate law is \(r=k[A]^2\).
If the concentration of \(A\) is doubled, the reaction rate will be
1 doubled
2 quadrupled
3 increased by 8 times
4 unchanged.
Explanation:
(b) \(r=k[A]^2\)
When concentration of \(A\) is doubled,
\(r=k[2 A]^2\)
\(r=4 k[A]^2\)
Karnataka CET 2016
CHEMISTRY(KCET)
285440
\(3 A \rightarrow 2 B\), rate of reaction, \(+\frac{d[B]}{d t}\) is equal to
