Question

1 of 15 A graph of this would produce the straightest line for a 2nd order reaction: natural log of concentration versus time O reciprocal of concentration versus time O concentration versus time O temperature versus time 2 of 15 Which statement is true about a reaction profile? o The energy of the reactants is always greater than that of the products. O The activation energy of the reverse reaction is not same as the forward. O The products must have more energy than the reactants. O The activated complex is more stable than the reactants.

Answer 1

1) Graph of this would produce the straightest line for 2 nd order reaction:

ANSWER : reciprocal of concentration versus time.

$\Rightarrow$ For second order reaction, we have integrated rate law equation 1/ [A] = k t + 1/ [A] ₀

Compare this equation with y = m x + c , we get y = 1/ [A] , x = t ,m = k and c = 1/ [A] ₀

We know that, if y and x are plotted , we get straight line making intercept on y axis.

Therefore, if we plot  1/ [A] against time t , we get straight line making intercept on y axis.

Slope = k= Rate constant of reaction 4 Intercept = \[A]. -+-+

2) Which statement is true about reaction profile.

a) The energy of reactants is always greater than that of the products.

FALSE

$\Rightarrow$ Some times energy of reactant is greater than product and some times it is less than products. Hence given statement is false.

b) The activation energy of reverse reaction is not same as forward reaction.

TRUE

Statement can be explained from figure shown below.

Activated compley Activated complex РЕ EROEP Ep> ER P Reaction coordinate Reaction wordinate F = Ea of forward reaction B-Ea of backward readion Eqt Eapis

C) Products must have more energy than reactants.

FALSE

D) The activated complex is more stable than reactant.

FALSE

Lower the energy more will be stability. If we see figure above , Activated complex always have higher energy than reactants and products. Hence, activated complex is always less stable than reactants and products.