Question

What is the value of the slope of a graph obtained by plotting the concentration of reactant \((A)_t\) versus time for zero order reaction?

• A. \(-k\)
• B. \(1/2k\)
• C. \(k/2.303\)
• D. \(2.303/k\)

Hint:

For zero order, [A] = [A]0 - kt, a straight line.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a zero order reaction the rate does not depend on the concentration of the reactant. The concentration falls by a fixed amount in each unit of time.

Step 2: Key Formula:
The integrated rate law is \([A]_t = [A]_0 - kt\).

Step 3: Compare with a straight line:
A straight line is \(y = mx + c\). Put \(y = [A]_t\) and \(x = t\). Then the slope is \(m = -k\) and the intercept is \([A]_0\).

Step 4: Check the options:
The other values do not come from this equation. \(k/2.303\) is a first order slope (for \(\log[A]\) against \(t\), with a negative sign), while \(1/2k\) and \(2.303/k\) are not slopes of any plot here. So the slope is \(-k\), option (A).

Final Answer:
The slope of [A] against t is -k. \[ \boxed{-k} \]