$$\bullet$$ For zero order reaction

r = k[A]⁰

r = k (constant)

hence, 'y' as 'rate' and 'x' as concentration will give desired graph.

$$\bullet$$ For first order reaction

$${t_{1/2}} = {{0.693} \over k}$$ (constant)

hence, 'y' as 't_1/2' and 'x' as concentration will given desired graph.

Arrhenius equation

$$k = A{e^{ - {E_a}/RT}}$$

$$\ln k = \ln A + \ln {e^{ - {E_a}/RT}}$$

$$\ln k = \ln A - {{{E_a}} \over R}\left( {{1 \over T}} \right)$$ .... (1)

Slope of $$\ln k\,$$ vs $${1 \over T}$$ curve,

$$m = - {{{E_a}} \over R}$$

$$ - 5 \times {10^3} = - {{{E_a}} \over R}$$

E_a = 5 $$\times$$ 10³ $$\times$$ 8.314 J/mol

= 41.57 $$\times$$ 10³ J/mol

$$ \simeq $$ 41.5 kJ/mol

The enthalpy of reaction is negative, – 4.2 kJ mol⁻¹ . The reaction is an exothermic reaction i.e. the energy of product B, is less than the energy of reactant A. So, the potential energy profile for the reaction is

$$k = \left( {{{2.303} \over t}} \right)$$ log$$\left( {{{{A_0}} \over A}} \right)$$ for first order reaction

$$4.606 \times {10^{ - 3}} = \left( {{{2.303} \over t}} \right)$$ log$$\left( {{2 \over {0.2}}} \right)$$

$$ \Rightarrow $$ t = 500 s