For a zero-order reaction, the plot of concentration vs time is linear with:

–ve slope and non-zero intercept

The correct answer is option 4. –ve slope and non-zero intercept.

Consider the general reaction:

\(A \rightarrow Products\)

If it is a reaction of zero order, then the rate of reaction will be

\(−\frac{d[A]}{dt} = k[A]^o\)

or, \(−\frac{d[A]}{dt}= k\)     \([Since, [A]^o = 1]\)

or, \(d[A]  = −kdt\)-----------------(i)

Integrating both sides, we get

\(\int{d[A]} = −k \int{dt}\)

or, \([A] = −kt + c\) ---------------(ii)

where, \(c\) is the constant of integration

At

\(t = 0, [A] = [A]_o\)

\(∴ [A]_o = c\)

Substituting this value in equation (i), we get

\([A] = −kt + [A]_o\)

or, \(kt = [A]_o − [A]\)

or, \(k = \frac{1}{t}{[A]_o − [A]}\) ---------------(iii)

This is the expression for rate constant for reactions of zero order.

From zero order reaction, we have

\([A] = −kt + [A]_o\)

As it is in the form of a straight line, \(y = mx + c\) , where, slope, \(m = − k\) and the intercept, \(c =[A]_o\). So the plot of concentration, [A] versus time, t will be