Question

What is the value of rate constant for first order reaction if slope for the graph of rate versus concentration is $2.5 \times 10^{-3}$?

• A. $2.5 \times 10^{-3}$ time$^{-1}$
• B. $5.0 \times 10^{-3}$ time$^{-1}$
• C. $7.5 \times 10^{-3}$ time$^{-1}$
• D. $1.25 \times 10^{-3}$ time$^{-1}$

Hint:

For a first-order reaction, plotting 'Rate' against '[Concentration]' yields a straight line passing through the origin, and its slope directly represents the rate constant 'k'.

Answer 1

The Correct Option is A

Step 1: Recall the rate law for a first-order reaction. For a first-order reaction, the rate is directly proportional to the concentration of one reactant. If we denote the reactant as A, the rate law is: \[ \text{Rate} = k[A] \] where: $k$ is the rate constant. $[A]$ is the concentration of the reactant.
Step 2: Understand the graph of Rate versus Concentration. If we plot \text{Rate} on the y-axis and $[A]$ (concentration) on the x-axis for a first-order reaction, the equation is in the form of a straight line, $y = mx + c$, where $y = \text{Rate}$ and $x = [A]$. In this case, the intercept $c$ is $0$, so the equation simplifies to $y = mx$. Comparing \text{Rate} $= k[A]$ with $y = mx$, we can see that the slope ($m$) of the graph of \text{Rate} versus $[A]$ is equal to the rate constant ($k$). \[ \text{Slope} = k \]
Step 3: Determine the value of the rate constant. The problem states that the slope for the graph of rate versus concentration is $2.5 \times 10^{-3}$. Since \text{Slope} $= k$, the rate constant $k$ is: \[ k = 2.5 \times 10^{-3} \text{ time}^{-1} \] The unit for a first-order rate constant is time$^{-1}$ (e.g., s$^{-1}$, min$^{-1}$).