Which one of the following given graphs represents the variation of rate constant (k) with temperatu

Q: Why does the question specifically mention an 'endothermic' reaction?

It's a classic distractor! Whether a reaction is endothermic or exothermic, its activation energy $E_a$ is always positive. Therefore, the fundamental shape of the $k$ vs $T$ graph—an exponential increase—remains exactly the same for both.

Q: Why isn't the graph a straight line if rate increases with temperature?

Because the relationship is not linear. The temperature $T$ sits in the denominator of an exponent in the Arrhenius equation ($k = A e^{-\frac{E_a}{RT}}$). This creates an exponential growth curve, not a linear one.

Q: Does the graph ever stop curving upwards?

Yes, mathematically it does! If you take the second derivative of the Arrhenius equation, you'll find an inflection point at $T = \frac{E_a}{2R}$. Beyond this extremely high temperature (often thousands of Kelvin), the curve becomes concave down and eventually asymptotes to $A$. However, at normal chemical temperatures, it is always concave up.

Q: What does the y-intercept of this graph represent?

At $T \to 0$, the exponent approaches $-\infty$, making $k \to 0$. So the graph starts from the origin (or very close to it). The pre-exponential factor $A$ is actually the horizontal asymptote as $T \to \infty$, not the y-intercept!

Q: Could the rate constant ever decrease with an increase in temperature?

For elementary reactions following Arrhenius behavior, no. Since $E_a$ is positive, an increase in $T$ will always increase $k$. However, for some complex multi-step reactions with negative apparent activation energies, the overall rate can decrease, but that's an exception, not the rule.

Question

Which one of the following given graphs represents the variation of rate constant ($k$) with temperature ($T$) for an endothermic reaction?

Sigma Percentile · Accepted Answer

The relationship between temperature and the rate of a chemical reaction is one of the most profound concepts in chemical kinetics. When we heat a reaction mixture, we are essentially pumping thermal energy into the system, causing molecules to move faster, collide more frequently, and, most importantly, collide with greater energy. But how exactly does this translate into the mathematical language of the rate constant, $k$? To answer this, we must turn to the legendary Arrhenius equation. ### The Master Equation: Arrhenius Theory The Arrhenius equation beautifully captures the temperature dependence of the rate constant: $$k = A e^{-\frac{E_a}{RT}}$$ Let us break down the anatomy of this equation. Here, $A$ is the pre-exponential factor (or frequency factor), representing the total number of collisions per second with the correct orientation. $E_a$ is the activation energy, the minimum energy barrier the molecules must overcome to react. $R$ is the universal gas constant, and $T$ is the absolute temperature in Kelvin. The most critical part of this equation is the exponential term, $e^{-\frac{E_a}{RT}}$. This term represents the fraction of molecules that possess enough kinetic energy to overcome the activation energy barrier. ### Analyzing the Mathematical Behavior Imagine you are slowly turning up the heat on a reaction. As the temperature $T$ increases, the denominator of the fraction $\frac{E_a}{RT}$ becomes larger. Consequently, the overall value of the fraction $\frac{E_a}{RT}$ becomes smaller. However, we must not forget the crucial negative sign in front of it! Because of this negative sign, as $\frac{E_a}{RT}$ gets smaller, the entire exponent $-\frac{E_a}{RT}$ becomes **less negative**. In mathematical terms, it is increasing towards zero. Since the exponent is increasing, the value of the exponential function $e^{-\frac{E_a}{RT}}$ must also increase. Therefore, the rate constant $k$ increases as temperature $T$ increases. This eliminates any graph that shows a decreasing trend. ### The Shape of the Curve: Why is it Concave Up? We know the graph goes up, but does it go up in a straight line? Because the temperature $T$ is trapped inside an exponent, the relationship is fundamentally non-linear. To understand the exact curvature, we can look at the first and second derivatives of $k$ with respect to $T$. The first derivative tells us the slope: $$\frac{dk}{dT} = A e^{-\frac{E_a}{RT}} \left( \frac{E_a}{RT^2} \right)$$ Since all terms here are positive, the slope is always positive, confirming that the graph is strictly increasing. Now, let us look at the second derivative to determine the concavity: $$\frac{d^2k}{dT^2} = A e^{-\frac{E_a}{RT}} \left( \frac{E_a}{RT^3} \right) \left[ \frac{E_a}{RT} - 2 \right]$$ For the graph to be concave up (curving upwards like a bowl), the second derivative must be positive. This happens when: $$\frac{E_a}{RT} - 2 > 0 \implies T < \frac{E_a}{2R}$$ For almost all chemical reactions, the activation energy $E_a$ is large enough (typically tens to hundreds of kJ/mol) that the value of $\frac{E_a}{2R}$ is extremely high—often several thousands of Kelvin. Since we conduct reactions at temperatures well below this threshold, the graph will always be **concave up**. It represents an exponential growth curve. ### The "Endothermic" Distractor The question specifically asks for the graph of an *endothermic* reaction. Is there a catch here? This is a classic trap designed to test your conceptual clarity! Whether a reaction is endothermic (absorbs heat, $\Delta H > 0$) or exothermic (releases heat, $\Delta H < 0$), the activation energy $E_a$ for the forward reaction is *always* a positive value. Because $E_a$ is always positive, the mathematical behavior of the Arrhenius equation remains identical. The rate constant $k$ will always increase exponentially with temperature, regardless of the reaction's thermodynamics. The endothermic label is merely a distractor. ### Final Conclusion By synthesizing our mathematical and physical understanding, we can confidently determine the correct graph. We are looking for a curve that starts near the origin and increases exponentially, curving upwards (concave up). Looking at the given options, the graph that perfectly matches this exponential upward curve is the one presented in option (c).

Animated Solution for Chemistry - Chemical Kinetics: Which one of the following given graphs represents the variation of rate constant ( $k$ ) with temperature ( $T$ ) for an endothermic reaction?

Select Answer:

Visualized Solution

$k = A e^{- \frac{E _{a}}{R T}}$

$Effect of Temperature$

$Exponential Growth$

$Endothermic vs Exothermic$

$The Inflection Point$

The Sigma Insight: Arrhenius Theory, Activation Energy and Collision Theory of Bimolecular Gaseous Reaction

The Master Equation

Analyzing the Mathematical Behavior

The Shape of the Curve

The "Endothermic" Distractor

Final Conclusion

Similar Questions

Consider the following plots of rate constant versus $\frac{1}{T}$ for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

Consider the given plot of enthalpy of the following reaction between $A$ and $B$ . $A + B ⟶ C + D$ Identify the incorrect statement.

The rate of a certain biochemical reaction at physiological temperature ( $T$ ) occurs $1 0^{6}$ times faster with enzyme than without. The change in the activation energy upon adding enzyme is :