The given plots represent the variation of the concentration of a reaction R with time for two diffe

Q: Why is the slope negative in both of these graphs?

The slope is negative because the concentration of the reactant $R$ is continuously decreasing as the reaction proceeds over time. The negative slope mathematically represents the rate constant $-k$.

Q: What exactly does the y-intercept represent in these kinetic graphs?

The y-intercept represents the state of the system at time $t = 0$. For the zero-order graph, it is the initial concentration $[R]_0$. For the first-order graph, it is the natural logarithm of the initial concentration, $\ln[R]_0$.

Q: How would the graph for a second-order reaction look?

For a second-order reaction, the integrated rate law is $\frac{1}{[R]} = kt + \frac{1}{[R]_0}$. Therefore, a plot of $\frac{1}{[R]}$ versus time $t$ would give a straight line, but unlike zero and first-order, it would have a **positive** slope equal to $k$.

Q: Can we calculate the rate constant $k$ directly from these graphs?

Yes, absolutely! In both of these specific graphs, the slope of the straight line is equal to $-k$. By calculating the slope ($\frac{\Delta y}{\Delta x}$) and taking its negative value, you get the rate constant.

Q: Why do we use natural log ($\ln$) instead of base-10 log ($\log$) in the first-order graph?

The integration of $\frac{1}{[R]}$ naturally yields the natural logarithm $\ln[R]$. While you can convert it to base-10 log using the relation $\ln(x) = 2.303 \log(x)$, the equation becomes $\log[R] = -\frac{k}{2.303}t + \log[R]_0$. A plot of $\log[R]$ vs $t$ is also a straight line, but its slope would be $-\frac{k}{2.303}$ instead of just $-k$.

Question

The given plots represent the variation of the concentration of a reaction $R$ with time for two different reactions (i) and (ii). The respective orders of the reactions are

Options:

(A) 1, 1

(B) 0, 2

(C) 0, 1

(D) 1, 0

Sigma Percentile · Accepted Answer

### Decoding Reaction Kinetics Through Graphs

Graphs are the visual language of chemical kinetics. They allow us to instantly identify the order of a reaction just by looking at which variables produce a straight line. In this problem, we are presented with two distinct straight-line graphs, both exhibiting a negative slope. The key to unlocking their meaning lies entirely in their y-axis labels.

### Analyzing the First Graph: $\ln[R]$ vs $t$

Let's start by recalling the fundamental rate law for a **first-order reaction**. In such a reaction, the rate of disappearance of the reactant $R$ is directly proportional to its concentration:

$$ 	ext{Rate} = -\frac{d[R]}{dt} = k[R] $$

To find how the concentration changes over time, we rearrange this differential equation and integrate it from time $t = 0$ (where concentration is $[R]_0$) to time $t$ (where concentration is $[R]$):

$$ \int_{[R]_0}^{[R]} \frac{d[R]}{[R]} = -k \int_0^t dt $$

Evaluating this integral yields the integrated rate law for a first-order reaction:

$$ \ln[R] - \ln[R]_0 = -kt $$

Rearranging this into the standard equation of a straight line, $y = mx + c$, we get:

$$ \ln[R] = -kt + \ln[R]_0 $$

If we plot $\ln[R]$ on the y-axis and time $t$ on the x-axis, the equation dictates that we will get a straight line. The slope ($m$) of this line will be $-k$, and the y-intercept ($c$) will be $\ln[R]_0$. This perfectly matches the first graph (i) given in the problem. Therefore, graph (i) represents a **first-order reaction**.

### Analyzing the Second Graph: $[R]$ vs $t$

Now, let's shift our focus to a **zero-order reaction**. For these reactions, the rate is entirely independent of the reactant's concentration:

$$ 	ext{Rate} = -\frac{d[R]}{dt} = k[R]^0 = k $$

Integrating this simpler differential equation over the same limits gives:

$$ \int_{[R]_0}^{[R]} d[R] = -k \int_0^t dt $$

$$ [R] - [R]_0 = -kt $$

Once again, we rearrange this into the $y = mx + c$ format:

$$ [R] = -kt + [R]_0 $$

This equation tells us that plotting the concentration $[R]$ directly against time $t$ will yield a straight line. The slope ($m$) is again $-k$, and the y-intercept ($c$) is the initial concentration $[R]_0$. This is an exact match for the second graph (ii). Thus, graph (ii) represents a **zero-order reaction**.

### Conclusion

By simply deriving the integrated rate laws and comparing them to the standard equation of a straight line, we have successfully decoded the graphs. Graph (i) corresponds to a first-order reaction, and graph (ii) corresponds to a zero-order reaction. The respective orders are 1 and 0, making option (d) the correct choice.

Animated Solution for Chemistry - Chemical Kinetics: The given plots represent the variation of the concentration of a reaction $R$ with time for two different reactions (i) and (ii). The respective orders of the reactions are

Select Answer:

Visualized Solution

Visual Anchor

First Order Rate Law

Integration Setup

Integrated Equation

Matching Graph (i)

Zero Order Rate Law

Integration Setup

Integrated Equation

Matching Graph (ii)

Final Conclusion

The Sigma Insight: Order and Molecularity

Decoding Reaction Kinetics Through Graphs

Analyzing the First Graph: $ln [R]$ vs $t$

Analyzing the Second Graph: $[R]$ vs $t$

Conclusion

Similar Questions

For the following graphs Choose from the options given below, the correct one regarding order of reaction is

Consider the following reactions $A \to P 1; B \to P 2; C \to P 3; D \to P 4$ , The order of the above reactions are $a, b, c$ and $d$ , respectively. The following graph is obtained when $lo g [rate]$ vs $lo g [conc.]$ are plotted :

For a reaction $A + 2 B ⟶ C$ , rate is given by $+ \frac{d [ C ]}{d t} = k [A] [B]$ , hence the order of the reaction is

Consider following two reactions, $A ⟶ Product; - \frac{d [ A ]}{d t} = k_{1} [A]^{0}$ $B ⟶ Product; - \frac{d [ B ]}{d t} = k_{2} [B]$ $k_{1}$ and $k_{2}$ are expressed in terms of molarity ( $mol L^{- 1}$ ) and time ( $s^{- 1}$ ) as

For the elementary reaction $M \to N$ , the rate of disappearance of $M$ increases by a factor of $8$ upon doubling the concentration of $M$ . The order of the reaction with respect to $M$ is

In a first order reaction, the concentration of the reactant, decreases from $0.8 M$ to $0.4 M$ in $15 min$ . The time taken for the concentration to change from $0.1 M$ to $0.025 M$ is

Select Answer:

Visualized Solution

Visual Anchor

First Order Rate Law

Integration Setup

Integrated Equation

Matching Graph (i)

Zero Order Rate Law

Integration Setup

Integrated Equation

Matching Graph (ii)

Final Conclusion

The Sigma Insight: Order and Molecularity

Decoding Reaction Kinetics Through Graphs

Analyzing the First Graph: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln[R] vs .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Analyzing the Second Graph: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [R] vs .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Conclusion

Similar Questions

For the following graphs Choose from the options given below, the correct one regarding order of reaction is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a reaction A+2B⟶C, rate is given by +dtd[C]​=k[A][B], hence the order of the reaction is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is

Analyzing the First Graph: $ln [R]$ vs $t$

Analyzing the Second Graph: $[R]$ vs $t$

For a reaction $A + 2 B ⟶ C$ , rate is given by $+ \frac{d [ C ]}{d t} = k [A] [B]$ , hence the order of the reaction is

In a first order reaction, the concentration of the reactant, decreases from $0.8 M$ to $0.4 M$ in $15 min$ . The time taken for the concentration to change from $0.1 M$ to $0.025 M$ is