A reaction is first-order with respect to A and second-order with respect to B. The concentration of B is increased three times. The new rate of the reaction would:

1.	Decrease 9 times	2.	Increase 9 times
3.	Increase 6 times	4.	Decrease 6 times

In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below:

The order of the reaction with respect to A and B would be:

For a reaction, 2A + B → C + D, the following observations were recorded:

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:

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 $4$ ${\mathrm{years}}^{-1}$. The value of half-life will be : 

1. 0.05 years

2. 0.17 years

3. 0.26 ${\mathrm{years}}^{-1}$

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 ${\mathrm{E}}_{\mathrm{a}}$ will be:

1. $76.64$ $J$ ${\mathrm{mol}}^{-1}$

2. $66.64$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

3. $76.64$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

4. $70.34$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

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:

$1. - 4.75$ $\times {10}^{-4}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$2.$ $5.75\times {10}^{-5}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$3.$ $$ $6.75\times {10}^{-6}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$4.$ $-6.75\times {10}^{-6}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$

The slope of the given below graph is -

 $1.$ $\frac{\mathrm{k}}{2.303}$
$$
$2.$ $-\frac{\mathrm{k}}{2.303}$
$$
$3.$ $-\frac{2.303}{\mathrm{k}}$
$$
$4.$ $\frac{2.303}{\mathrm{k}}$