For zero order reaction, A → ,a graph of rate vs time has slope equal to:( where k = rate of reaction) A. k B. – k C. zero D. – 2.303 k
The Correct Answer and Explanation is:
ChatGPT said:
Correct Answer: C. Zero
Explanation
In a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. This means the rate remains constant throughout the reaction as long as some reactant is still present.
The general form of a zero-order reaction is:A→Products\text{A} \rightarrow \text{Products}A→Products
For a zero-order reaction, the rate law is:Rate=−d[A]dt=k\text{Rate} = -\frac{d[\text{A}]}{dt} = kRate=−dtd[A]=k
Here, kkk is the rate constant, and the negative sign indicates the concentration of A is decreasing.
Now, if we graph rate vs. time for a zero-order reaction:
- The rate =k= k=k, which is a constant.
- This means no matter how much time passes, the rate doesn’t change as long as A is present.
So, plotting rate on the y-axis and time on the x-axis results in a horizontal line. A horizontal line has a slope of zero.
Therefore, the slope of the rate vs. time graph is zero, which corresponds to:C. Zero\boxed{\text{C. Zero}}C. Zero
Additional Context:
- For zero-order reactions, the integrated rate law is: [A]=[A]0−kt[A] = [A]_0 – kt[A]=[A]0−kt This tells us how the concentration of A decreases linearly with time.
- However, the rate itself does not change, because it’s not dependent on [A].
- This behavior is typical in reactions where a catalyst surface becomes saturated or in enzyme kinetics under conditions of maximum saturation.
- In contrast:
- First-order reactions have rates that decrease exponentially over time.
- Second-order reactions show even faster decreases in rate with time.
But in zero-order, the unchanging rate gives a flat line in a rate vs. time plot — hence a slope of zero.
