In conversion of limestone to lime, CaCO3(s) - CaO(s) + CO2(g) the values of ΔH° and ΔS° are +179.1

Q: Why do we set $\Delta G = 0$ if we want the reaction to be spontaneous?

Setting $\Delta G = 0$ gives us the equilibrium temperature. This is the exact boundary line. Once we find this threshold temperature, we know that any temperature higher than this will make $\Delta G$ negative, thus making the reaction spontaneous.

Q: Why does increasing the temperature make this specific reaction spontaneous?

Because both $\Delta H$ and $\Delta S$ are positive. In the equation $\Delta G = \Delta H - T\Delta S$, the $T\Delta S$ term is subtracted. At high temperatures, $T\Delta S$ becomes larger than $\Delta H$, resulting in a negative $\Delta G$.

Q: What would happen if $\Delta S$ was negative?

If $\Delta S$ was negative, the $-T\Delta S$ term would become positive. Since $\Delta H$ is already positive, $\Delta G$ would be positive at all temperatures, meaning the reaction would never be spontaneous at any temperature.

Q: I divided 179.1 by 160.2 and got 1.11. Why is my answer wrong?

You fell into the classic unit trap! $\Delta H$ is given in kilojoules (kJ), while $\Delta S$ is in joules (J). You must multiply $\Delta H$ by 1000 to convert it to joules before dividing.

Q: Does the pressure of 1 bar mentioned in the question affect our calculation?

The 1 bar pressure simply indicates that the given $\Delta H^\circ$ and $\Delta S^\circ$ values are standard state values. It assures us that we can directly use the standard Gibbs free energy equation without needing to adjust for non-standard pressures.

Question

In conversion of limestone to lime,
$	ext{CaCO}_3(s) \longrightarrow 	ext{CaO}(s) + 	ext{CO}_2(g)$
the values of $\Delta H^\circ$ and $\Delta S^\circ$ are $+179.1 	ext{ kJ mol}^{-1}$ and $160.2 	ext{ J /K}$, respectively at 298 K and 1 bar. Assuming that $\Delta H^\circ$ and $\Delta S^\circ$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

Options:

(A) 1008 K

(B) 1200 K

(C) 845 K

(D) 1118 K

Sigma Percentile · Accepted Answer

### The Thermodynamics of Limestone

Imagine you have a piece of limestone, chemically known as calcium carbonate ($	ext{CaCO}_3$), and you want to convert it into quicklime ($	ext{CaO}$). This is a classic thermal decomposition reaction. But the fundamental question is: will this reaction happen on its own?

To answer that, we need to dive into the thermodynamics of the process. We are looking for the conditions under which this reaction becomes **spontaneous**.

### The Master Equation: Gibbs Free Energy

The master key to understanding spontaneity is the Gibbs Free Energy equation:

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

For any reaction to be spontaneous, the change in Gibbs free energy ($\Delta G$) must be strictly less than zero ($\Delta G < 0$). Let's analyze the components of this equation for our specific reaction.

Since the decomposition of limestone absorbs heat, it is an endothermic process, meaning the enthalpy change ($\Delta H$) is positive. Furthermore, a solid is decomposing to produce a solid and a gas ($	ext{CO}_2$). The creation of a gas significantly increases the randomness or disorder of the system, making the entropy change ($\Delta S$) positive as well.

### The Tipping Point

Because both $\Delta H$ and $\Delta S$ are positive, the sign of $\Delta G$ depends entirely on the temperature, $T$. At low temperatures, the positive $\Delta H$ dominates, making $\Delta G$ positive (non-spontaneous). However, as the temperature increases, the $T\Delta S$ term grows larger.

The exact moment the reaction switches from non-spontaneous to spontaneous is the equilibrium point, where $\Delta G$ is exactly zero.

$$0 = \Delta H - T\Delta S$$

We can rearrange this equation to find the threshold temperature, $T$:

$$T = \frac{\Delta H}{\Delta S}$$

Any temperature strictly above this threshold will make the $T\Delta S$ term larger than $\Delta H$, pushing $\Delta G$ into the negative territory and driving the reaction forward.

### The Unit Trap

Now, let's substitute the given values. But wait, there is a catch here! This is a favorite trap in competitive exams like JEE.

We are given:
$\Delta H = +179.1 	ext{ kJ mol}^{-1}$
$\Delta S = 160.2 	ext{ J K}^{-1} 	ext{mol}^{-1}$

Notice the units? $\Delta H$ is in kilojoules, while $\Delta S$ is in joules. We must convert $\Delta H$ to joules before dividing.

$$\Delta H = 179.1 	imes 10^3 	ext{ J mol}^{-1}$$

### The Final Verdict

Once the units match, we plug them into our threshold temperature formula:

$$T = \frac{179.1 	imes 10^3}{160.2}$$

$$T = \frac{179100}{160.2} \approx 1117.97 	ext{ K}$$

Rounding it off to the nearest integer, we get $1118 	ext{ K}$. This is our equilibrium temperature. For the reaction to be spontaneous, the temperature must be strictly greater than this threshold.

Therefore, the temperature above which the conversion of limestone to lime will be spontaneous is **$1118 	ext{ K}$**.

Select Answer:

Visualized Solution

\text{Understanding Spontaneity}

\text{Gibbs Free Energy Equation}

\text{Finding the Threshold Temperature}

\text{Unit Conversion and Substitution}

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Entropy and Free Energy

The Thermodynamics of Limestone

The Master Equation

The Tipping Point

The Unit Trap

The Final Verdict

Similar Questions

The reaction, $MgO (s) + C (s) ⟶ Mg (s) + CO (g)$ , for which $Δ_{r} H^{\circ} = + 491.1 kJ mol^{- 1}$ and $Δ_{r} S^{\circ} = 198.0 J K^{- 1} mol^{- 1}$ , is not feasible at 298 K. Temperature above which reaction will be feasible is

A process has $Δ H = 200 J mol^{- 1}$ and $Δ S = 40 J K^{- 1} mol^{- 1}$ . Out of the values given below, choose the minimum temperature above which the process will be spontaneous

For a particular reversible reaction, at temperature $T$ , $Δ H$ and $Δ S$ were found to be both +ve. If $T_{e}$ is the temperature at equilibrium, the reaction would be spontaneous when

A process will be spontaneous at all temperature if

A reaction is non-spontaneous at the freezing point of water but is spontaneous at the boiling point of water, then

Standard entropy of $X_{2}, Y_{2}$ and $X Y_{3}$ are $60$ , $40$ and $50 JK^{- 1} mol^{- 1}$ , respectively. For the reaction, $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⟶ X Y_{3}, Δ H = - 30 kJ$ , to be at equilibrium, the temperature will be

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 ...... $JK^{- 1} mol^{- 1}$ .