For a general reaction A $\to$ B, the plot of the concentration of A vs. time is given in the figure.
 

The slope of the curve will be:

1.	-k	2.	-k/2
3.	-k2	4.	-k/3

The decomposition of hydrocarbons follows the equation: k = (4.5 × 10¹¹s^–1) e^-28000K/T

The activation energy (E_a) for the reaction would be:

The rate constant for a first-order reaction is $4.606\times {10}^{-3}$ $s^{-1}$. The time required to reduce 2.0 g of the reactant to 0.2 g will be:

The role of a catalyst is to change: 

1. Gibbs energy of the reaction

2. Enthalpy of reaction

3. The activation energy of the reaction 

4. Equilibrium constant

The correct statement based on the graph below is:

Consider the first-order gas-phase decomposition reaction given below.

A(g) → B(g) + C(g)

The initial pressure of the system before the decomposition of A was ${\mathrm{P}}_{\mathrm{i}}$. After the lapse of time t, the total pressure of the system increased by X units and became ${\mathrm{P}}_{\mathrm{t}}$. The rate constant k for the reaction is:

1. $\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{P_{\mathrm{i}}}{P_{\mathrm{i}}-\mathrm{x}}$

2. $\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{P_{\mathrm{i}}}{2P_{\mathrm{i}}-P_{\mathrm{t}}}$

3. $\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{P_{\mathrm{i}}}{2P_{\mathrm{i}}+P_{\mathrm{t}}}$

4. $\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{P_{\mathrm{i}}}{P_{\mathrm{i}}+\mathrm{x}}$

The correct graphical representation of relation between ln k and 1/T is:

1.		2.
3.		4.

True statement among the following is:

The correct expression for the rate of reaction given below is:
\(5 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{Br}_2(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

The correct graphical representation of first-order reaction is:

(a)		(b)
(c)		(d)