At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol^-1) for hydrogen (E_H)

Q: Why do $H_2$ and $D_2$ have the same electronic potential energy curve?

Because isotopes have the same number of protons and electrons, their electrostatic interactions and chemical bonding are identical. The potential energy curve is determined purely by these electronic interactions.

Q: What exactly is Zero-Point Energy?

It's the lowest possible energy that a quantum mechanical physical system may have. Unlike in classical mechanics, quantum systems constantly fluctuate and vibrate even in their lowest energy state.

Q: Why does $D_2$ have a lower ZPE than $H_2$?

ZPE is proportional to the vibrational frequency, which is inversely proportional to the square root of the reduced mass. Since Deuterium is heavier than Hydrogen, $D_2$ has a lower frequency and thus a lower ZPE.

Q: Do I need to memorize these exact bond energy values for JEE?

Not necessarily the exact decimal values, but you must know the trend ($E_D > E_H$) and knowing that the approximate difference is around $7.5 \text{ kJ mol}^{-1}$ is highly beneficial for specific questions like this.

Q: What is the Kinetic Isotope Effect?

It's the change in the rate of a chemical reaction when one of the atoms in the reactants is replaced by one of its isotopes. Because C-D bonds are stronger than C-H bonds, reactions that involve breaking a C-D bond are noticeably slower.

Question

At $298.2	ext{ K}$ the relationship between enthalpy of bond dissociation (in $	ext{kJ mol}^{-1}$) for hydrogen ($E_H$) and its isotope, deuterium ($E_D$), is best described by

Options:

(A) $E_H = \frac{1}{2} E_D$

(B) $E_H = E_D$

(C) $E_H \simeq E_D - 7.5$

(D) $E_H = 2 E_D$

Sigma Percentile · Accepted Answer

### The Shared Potential Well

When we compare the bond dissociation energies of hydrogen ($H_2$) and its heavier isotope, deuterium ($D_2$), we must first look at their electronic structure. Because isotopes possess the exact same number of protons and electrons, their electrostatic interactions are identical.

This means that if we plot the potential energy of the molecule against the internuclear distance, both $H_2$ and $D_2$ share the exact same potential energy curve. Classically, one might assume that breaking their bonds would require the exact same amount of energy. However, the quantum world has a surprise for us.

### The Quantum Jiggle: Zero-Point Energy

Quantum mechanics dictates that a molecule can never be completely at rest. Even at absolute zero, a diatomic molecule continues to vibrate. This residual motion is known as **Zero-Point Energy (ZPE)**.

The vibrational frequency ($
u$) of a diatomic molecule is inversely proportional to the square root of its reduced mass ($\mu$). Because a deuterium atom is twice as massive as a hydrogen atom, the $D_2$ molecule has a larger reduced mass. Consequently, it vibrates more sluggishly, resulting in a lower vibrational frequency and, crucially, a **lower Zero-Point Energy**.

### Climbing Out of the Well

Bond dissociation energy is defined as the energy required to take the molecule from its lowest possible energy state (its ZPE) all the way up to the dissociation limit, where the atoms are completely separated ($E=0$).

Because $D_2$ has a lower ZPE, it sits deeper within the potential energy well compared to $H_2$. Therefore, $D_2$ has a longer "climb" to reach the dissociation limit. This is why the bond dissociation energy of deuterium ($E_D$) is strictly greater than that of hydrogen ($E_H$).

### The Final Calculation

Experimentally, the bond dissociation energy for hydrogen is $E_H = 435.88 	ext{ kJ mol}^{-1}$, while for deuterium it is $E_D = 443.35 	ext{ kJ mol}^{-1}$.

If we calculate the difference between these two values:

$$E_D - E_H = 443.35 - 435.88 = 7.47 	ext{ kJ mol}^{-1}$$

Rounding this off, we find that the difference is approximately $7.5 	ext{ kJ mol}^{-1}$. Rearranging this gives us the final relationship:

**$$E_H \simeq E_D - 7.5$$**

This subtle quantum mechanical difference is the driving force behind the **Kinetic Isotope Effect**, a phenomenon where replacing hydrogen with deuterium can significantly slow down the rate of a chemical reaction!

Animated Solution for Chemistry - Chemical Thermodynamics: At $298.2 K$ the relationship between enthalpy of bond dissociation (in $kJ mol^{- 1}$ ) for hydrogen ( $E_{H}$ ) and its isotope, deuterium ( $E_{D}$ ), is best described by

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Visualized Solution

$H_{2} vs D_{2} Potential Energy$

$Zero-Point Energy (ZPE)$

$Bond Dissociation Energy$

$E_{D} > E_{H}$

$Final Relationship$

$Kinetic Isotope Effect$

The Sigma Insight: Enthalpy and Hess's Law

The Shared Potential Well

The Quantum Jiggle

Climbing Out of the Well

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​ vs D2​ Potential Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Zero-Point Energy (ZPE)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bond Dissociation Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ED​>EH​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Relationship

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinetic Isotope Effect