For the reaction; A(l) - 2B(g)
ΔU = 2.1 kcal, ΔS = 20 cal K^-1 at 300 K.
Hence, ΔG in kcal is .....

Q: Why do we ignore the liquid reactant when calculating $\Delta n_g$?

The volume occupied by liquids and solids is negligible compared to the volume occupied by gases. Therefore, the pressure-volume work ($P\Delta V$) is almost entirely due to the expansion or contraction of gases, which is why we only count gaseous moles in $\Delta n_g$.

Q: Why is the value of $R$ taken as 2 instead of 8.314?

The value of the universal gas constant $R$ depends on the units being used. When working with Joules, $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$. However, when working with calories, $R \approx 2 \text{ cal K}^{-1} \text{ mol}^{-1}$.

Q: Why did we divide $\Delta S$ and $R$ by 1000?

The internal energy $\Delta U$ was given in kilocalories (kcal), while $\Delta S$ and $R$ were in calories (cal). To add or subtract these terms, they must be in the exact same unit. Dividing calories by 1000 converts them into kilocalories.

Q: What does a negative $\Delta G$ signify physically?

A negative $\Delta G$ means that the system has free energy available to do useful work, and the process will occur spontaneously in the forward direction without any continuous external energy input.

Q: Can we calculate $\Delta G$ directly without finding $\Delta H$ first?

Yes, you can combine the equations into a single step: $\Delta G = \Delta U + \Delta n_g RT - T\Delta S$. This is exactly what we did mathematically, but breaking it down into $\Delta H$ first helps prevent calculation errors and builds better conceptual intuition.

Question

For the reaction; $A(l) \longrightarrow 2B(g)$
$\Delta U = 2.1 	ext{ kcal}, \Delta S = 20 	ext{ cal K}^{-1}$ at $300 	ext{ K}$.
Hence, $\Delta G$ in kcal is ......

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a chemical reaction in a closed beaker. We have a liquid reactant, $A(l)$, which is steadily converting into two moles of a gaseous product, $2B(g)$. The temperature of the system is held constant at $300 	ext{ K}$.

Whenever a liquid transforms into a gas, the molecules break free from their intermolecular bonds and spread out. This massive increase in randomness means the entropy of the system is increasing. The problem provides us with the change in internal energy, $\Delta U = 2.1 	ext{ kcal}$, and the change in entropy, $\Delta S = 20 	ext{ cal K}^{-1}$. Our ultimate goal is to determine the change in Gibbs free energy, $\Delta G$, which will tell us if this reaction is spontaneous.

### The Master Equation

To find the Gibbs free energy change, we rely on the master equation of thermodynamics:

$$\Delta G = \Delta H - T\Delta S$$

However, there is a slight hurdle. We are given $\Delta U$, not the enthalpy change $\Delta H$. We must first bridge this gap using the relationship between enthalpy and internal energy for reactions involving gases:

$$\Delta H = \Delta U + \Delta n_g RT$$

Here, $\Delta n_g$ represents the change in the number of moles of gas. Looking at our balanced equation, $A(l) \longrightarrow 2B(g)$, the product side has $2$ moles of gas, while the reactant side has $0$ (since it is a liquid). Therefore, $\Delta n_g = 2 - 0 = 2 	ext{ mol}$.

### The Unit Trap

Before we rush into substituting numbers, we must pause and inspect our units. This is where many students fall into a classic trap! The internal energy $\Delta U$ is given in **kilocalories** (kcal), but the entropy $\Delta S$ and the universal gas constant $R$ are typically in **calories** (cal).

To maintain strict unit consistency, we must convert $\Delta S$ and $R$ into kilocalories by dividing them by $1000$:

$$\Delta S = \frac{20}{1000} 	ext{ kcal K}^{-1}$$

$$R = \frac{2}{1000} 	ext{ kcal K}^{-1} 	ext{ mol}^{-1}$$

### Final Calculation

Now, we are ready to execute the math. First, let's calculate the enthalpy change $\Delta H$:

$$\Delta H = 2.1 + \left(2 	imes \frac{2}{1000} 	imes 300ight)$$

$$\Delta H = 2.1 + \frac{1200}{1000} = 2.1 + 1.2 = 3.3 	ext{ kcal}$$

Next, we evaluate the entropy term, $T\Delta S$, which represents the energy unavailable to do useful work due to the system's randomness:

$$T\Delta S = 300 	imes \frac{20}{1000} = \frac{6000}{1000} = 6 	ext{ kcal}$$

Finally, we bring everything together into the Gibbs free energy equation:

$$\Delta G = 3.3 - 6 = -2.70 	ext{ kcal}$$

Because $\Delta G$ is negative, we can confidently conclude that this reaction is **spontaneous** at $300 	ext{ K}$. The massive increase in entropy (the $T\Delta S$ term) overpowers the endothermic nature of the reaction (the positive $\Delta H$), driving the process forward!

Animated Solution for Chemistry - Chemical Thermodynamics: For the reaction; $A (l) ⟶ 2 B (g)$ $Δ U = 2.1 kcal, Δ S = 20 cal K^{- 1}$ at $300 K$ . Hence, $Δ G$ in kcal is ......

Enter Numerical Value:

Visualized Solution

\text{Reaction Setup}

\Delta G = \Delta H - T\Delta S

\Delta n_g = n_{p(g)} - n_{r(g)}

\text{Unit Consistency}

\Delta H = \Delta U + \Delta n_g RT

T\Delta S

\Delta G = \Delta H - T\Delta S

\text{Spontaneity and Temperature}

The Sigma Insight: Entropy and Free Energy

Analyzing the Setup

The Master Equation

The Unit Trap

Final Calculation

Similar Questions

For a dimerisation reaction $2 A (g) ⟶ A_{2} (g)$ , at $298 K$ , $Δ U^{⊖} = - 20 kJ mol^{- 1}$ , $Δ S^{⊖} = - 30 JK^{- 1} mol^{- 1}$ , then the $Δ G^{⊖}$ will be ......... $J$ .

The standard reaction Gibbs energy for a chemical reaction at an absolute temperature $T$ is given by, $Δ_{r} G^{\circ} = A - B T$ Where $A$ and $B$ are non-zero constants. Which of the following is true about this reaction?

For a given chemical reaction, $A ⟶ B$ at $300 K$ the free energy change is $- 49.4 kJ mol^{- 1}$ and the enthalpy of reaction is $51.4 kJ mol^{- 1}$ . The entropy change of the reaction is ...... $JK^{- 1} mol^{- 1}$ .

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Animated Solution for Chemistry - Chemical Thermodynamics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the reaction; A(l)⟶2B(g) ΔU=2.1 kcal,ΔS=20 cal K−1 at 300 K. Hence, ΔG in kcal is ......

Enter Numerical Value:

Visualized Solution

\text{Reaction Setup}

\Delta G = \Delta H - T\Delta S

\Delta n_g = n_{p(g)} - n_{r(g)}

\text{Unit Consistency}

\Delta H = \Delta U + \Delta n_g RT

T\Delta S

\Delta G = \Delta H - T\Delta S

\text{Spontaneity and Temperature}

The Sigma Insight: Entropy and Free Energy

Analyzing the Setup

The Master Equation

The Unit Trap

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a dimerisation reaction 2A(g)⟶A2​(g), at 298 K, ΔU⊖=−20 kJ mol−1, ΔS⊖=−30 JK−1 mol−1, then the ΔG⊖ will be ......... J.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the equilibrium, 2H2​O⇌H3​O++OH−, the value of ΔG∘ at 298 K is approximately

Animated Solution for Chemistry - Chemical Thermodynamics: For the reaction; $A (l) ⟶ 2 B (g)$ $Δ U = 2.1 kcal, Δ S = 20 cal K^{- 1}$ at $300 K$ . Hence, $Δ G$ in kcal is ......

For a dimerisation reaction $2 A (g) ⟶ A_{2} (g)$ , at $298 K$ , $Δ U^{⊖} = - 20 kJ mol^{- 1}$ , $Δ S^{⊖} = - 30 JK^{- 1} mol^{- 1}$ , then the $Δ G^{⊖}$ will be ......... $J$ .

For the equilibrium, $2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ , the value of $Δ G^{\circ}$ at $298 K$ is approximately