Note: In the following questions match the items of Column I with appropriate item given in Column II.

Match the graph given in Column I with the order of reaction given in Column II. More than one item in Column I may link to the same item of Column II.

Column I Column II

(i)

(ii) (a) first order

(iii) (b) zero order

(iv)

(i)→(a),(ii)→(b),(iii)→(b),(iv)→(a)

(i) →(a),(iv) →(a)

For the first order of a reaction, the rate is directly proportional to the molar concentration of the reactant.

Thus,

Rate=-d[R]/dt=kR⟹-(d[R])/R=kdt

Thus, The above equation gives graph (i) for a first-order reaction.

Integrating both sides from t=0 to t=t at concentration R=R0 to R=R respectively.

We get,

log⁡〖 [R]〗=-kt/2.303+log⁡[R_0]

Where K is the rate constant, R is the molar concentration of the reactant at time t and R0 is the molar concentration of a reactant at time t=0.

Above equation represents the equation of a line i.e. y=mx+c, for y=log [R] and x=t.

Slope is -k/2.303.

Thus, (iv) →(a).

(ii) →(b),(iii) →(b)

For a zero-order reaction, Rate is only depending on rate constant.

Rate=k〖[R]〗^0

Where k is the rate constant and R is the molar concentration of a reactant.

Thus, the rate is constant with respect to the concentration of a reactant. Giving, graph (ii) for a zero-order reaction.

From the above equation, we get,

-(d[R])/dt=k

Integrating both sides from t=0 to t=t at concentration R0 to R respectively. We get,

[R]=-kt+[R_0]

Where R is the molar concentration of a reactant at t=t, R0 is the molar concentration of a reactant at t=0.

The above equation represents the equation of a line i.e. y=mx+c.

Thus, for a concentration vs time graph for a zero-order reaction will be a straight line with slope equals -k.

Thus, the graph (iii) is for a zero-order reaction.