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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Chemistry Test1.
This reaction follows first order kinetics. The rate constant at particular temperature is 2.303 x 10-2 hour-1. The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes? (log 2 = 0.3010)
0.125 M
0.215 M
0.25 x 2.303 M
0.05 M
2.
After 2 hours, a radioactive substance becomes \(\left( \frac { 1 }{ 16 } \right) ^{ th }\) of original amount Then the half life (in min) is _______.
60 minutes
120 minutes
30 minutes
15 minutes
3.
The correct difference between first and second order reactions is that________.
A first order reaction can be catalysed; a second order reaction cannot be catalysed.
The half life of a first order reaction does not depend on [A0]; the half life of a second order reaction does depend on [A0].
The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations.
4.
If 75% of a first order reaction was completed in 60 minutes, 50% of the same reaction under the same conditions would be completed in_______.
20 minutes
30 minutes
35 minutes
75 minutes
5.
In a homogeneous reaction A⟶B+C+D, the initial pressure was P0 and after time t it was P expression for rate constant in terms of P0, P and t will be _____.
\(k=\left( \frac { 2.303 }{ t } \right) \log\left( \frac { 2{ P }_{ 0 } }{ { 3P }_{ 0 }-P } \right) \)
\(k=\left( \frac { 2.303 }{ t } \right) \log\left( \frac { { 2P }_{ 0 } }{ { P }_{ 0 }-P } \right) \)
\(k=\left( \frac { 2.303 }{ t } \right) \log\left( \frac { 3{ P }_{ 0 }-P }{ 2P_{ 0 } } \right) \)
\(k=\left( \frac { 2.303 }{ t } \right) \log\left( \frac { 2{ P }_{ 0 } }{ { 3P }_{ 0 }-2P } \right) \)
1.
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
2.303 x 10-2 hour-1 = \(\frac { 2.303 }{ 1806 min } log\frac { \left[ { 0.25 }_{ } \right] }{ \left[ A \right] } \)
\(=\left(\frac{2.303 \times 10^{-2} hour^{-1 }\times 1806 min}{2.303}\right) = \log \left(\frac{0.25}{A}\right) \)
\(=\left(\frac{ 1806 \times 10 ^{-2}}{60}\right) = \log \left(\frac{0.25}{A}\right) \)
\(= 0.301 = \log \left(\frac{0.25}{A}\right) \)
\(=\log2 = \log \left(\frac{0.25}{A}\right) \)
\(2 = \log \left(\frac{0.25}{A}\right) \)
\([A] = \log \left(\frac{0.25}{2}\right) = 0.125 M\)
2.
3.
\({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
For a second order reaction
\(\mathrm{t}_{1 / 2}=\frac{2^{\mathrm{n}-1}-1}{(\mathrm{n}-1) \mathrm{k} \cdot\left[\mathrm{A}_{0}\right]^{\mathrm{n}-1}}\)
n = 2
\(\mathrm{t}_{1 / 2}=\frac{2^{\mathrm{2}-1}-1}{(\mathrm{n}-1) \mathrm{k} \cdot\left[\mathrm{A}_{0}\right]^{\mathrm{2}-1}}\)
\(\mathrm{t}_{1 / 2} =\frac{1}{ \mathrm{k}\left[\mathrm{A}_{0}\right]}\)
4.
t75% = 2t50%
t50% = (t75%/2) = (60/2) = 30 minutes
5.
| A | ⟶ | B | C | D | |
| Initial rate (M s-1) | a | 0 | 0 | 0 | |
| Reaction number | x | - | - | - | - |
| After time t | (a - x) | x | x | x | |
| Total number of moles | = (a + 2x) |
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