For a reaction, 2A + B → C + D, the following observations were recorded:
Experiment | [A]/mol L–1 | [B]/mol L–1 | Initial rate of formation of D/mol L–1 min–1 |
I | 0.1 | 0.1 | 6.0 × 10–3 |
II | 0.3 | 0.2 | 7.2 × 10–2 |
III | 0.3 | 0.4 | 2.88 × 10–1 |
IV | 0.4 | 0.1 | 2.40 × 10–2 |
The rate law applicable to the above mentioned reaction would be:
1. Rate = k[A]2[B]3
2. Rate = k[A][B]2
3. Rate = k[A]2[B]
4. Rate = k[A][B]
Given the following observations:
Experiment | [A] / mol L–1 | [B] / mol L–1 | Initial rate / mol L–1 min–1 |
I | 0.1 | 0.1 | 2.0 × 10–2 |
II | X | 0.2 | 4.0 × 10–2 |
III | 0.4 | 0.4 | Y |
The reaction between A and B is first-order with respect to A and zero-order with respect to B. The values of X and Y are, respectively:
1. X = 0.2 \(mol\) \(L^{- 1}\); Y = \(\) \(0 . 08\) \(mol\) \(L^{- 1} \left(min\right)^{- 1}\)
2. X = 0.02 \(mol\) \(L^{- 1}\); Y = \(\) \(0 . 08\) \(mol\) \(L^{- 1} \left(min\right)^{- 1}\)
3. X = 0.01 \(mol\) \(L^{- 1}\); Y = \(\) \(0 . 8\) \(mol\) \(L^{- 1} \left(min\right)^{- 1}\)
4. X = 0.2 \(mol\) \(L^{- 1}\); Y = \(\) \(0 . 8\) \(mol\) \(L^{- 1} \left(min\right)^{- 1}\)
The rate constant of a radioactive substance is . The value of half-life will be :
1. 0.05 years
2. 0.17 years
3. 0.26
4. 1.6 years
During a nuclear explosion, one of the products is 90Sr with a half-life of 28.1 years. If 1µg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, the amount of 90Sr that will remain after 10 years in the now grown up child would be -
(Given ,antilog(0.108)=1.28)
1. 0.227 µg
2. 0.781 µg
3. 7.81 µg
4. 2.27 µg
A first-order reaction's 10 percent completion time at 298 K is the same as its 25 percent completion time at 308 K. The value of will be:
1.
2.
3.
4.
Based on the graph below, the average rate of reaction will be:
1. \(\frac{[R_{2}]-[R_{1}]}{t_{2}-t_{1}}\)
2. \(-(\frac{[R_{2}]-[R_{1}]}{t_{2}-t_{1}})\)
3. \(\frac{[R_{2}]}{t_{2}}\)
4. \(-(\frac{[R_{1}]-[R_{2}]}{t_{2}-t_{1}})\)
Consider the following graph:
The instantaneous rate of reaction at t = 600 sec will be:
The nature of the reaction represented in the following graph is:
1. | Endothermic reaction |
2. | Exothermic reaction |
3. | Both endothermic and exothermic reactions are represented by the same graph. |
4. | None of the above |
The graph between lnK and 1/T is given below:
The value of activation energy would be:
1. | \(207.8\ \mathrm{KJ} / \mathrm{mol} \) | 2. | \(- 207.8\ \mathrm{KJ} / \mathrm{mol} \) |
3. | \(210.8\ \mathrm{KJ} / \mathrm{mol} \) | 4. | \(-210.8\ \mathrm{KJ} / \mathrm{mol} \) |