Question

Steam undergoes decomposition at high temperature as per the reaction : \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}^{\circ}=200 \mathrm{~kJ} \mathrm{~mol}^{-1} \Delta \mathrm{S}^{\circ}=40 \mathrm{~kJ} \mathrm{~mol}^{-1} .\) The temperature at which equilibrium constant is unity is : (1) 3000 Kelvin (2) 5000 Kelvin (3) 5333 Kelvin (4) 5 Kelvin

Short answer

The temperature is 5 K.

Step-by-step solution

Understand Gibbs Free Energy equation

The Gibbs free energy change (\triangle G) can be related to the equilibrium constant (K) with the equation: \[ \triangle G^\theta = -RT \text{ln}K \] When K = 1, \text{ln}K = 0, hence: \[ \triangle G^\theta = 0 \]

Relationship between \triangle G^\theta, \triangle H^\theta, and \triangle S^\theta

Recall that: \[ \triangle G^\theta = \triangle H^\theta - T \triangle S^\theta \] With \triangle G^\theta = 0 (from Step 1), the equation simplifies to: \[ 0 = \triangle H^\theta - T \triangle S^\theta \] Rearrange the equation to solve for T: \[ T = \frac{\triangle H^\theta}{\triangle S^\theta} \]

Substituting given values

Substitute the given values \triangle H^\theta = 200 \text{ kJ/mol} and \triangle S^\theta = 40 \text{ kJ/mol} into the equation: \[ T = \frac{200 \text{ kJ/mol}}{40 \text{ kJ/mol}} \]

Calculate the temperature

Perform the division to find the temperature: \[ T = \frac{200}{40} = 5 \text{ K} \]

Choose the correct option

Among the given options, the calculated temperature matches option (4).

Key concepts

Gibbs Free Energy

Gibbs free energy (\triangle G) is a crucial concept in thermodynamics. It helps predict the spontaneity of a reaction. A reaction goes in a spontaneous direction if \triangle G is negative. The relationship between Gibbs free energy and the equilibrium constant is given by the equation:
\[ \triangle G^\theta = -RT \text{ln}K \]
Here, R is the universal gas constant, and T is the temperature in Kelvin. When the equilibrium constant K equals 1, \text{ln}K becomes 0, making \[ \triangle G^\theta = 0. \] This implies that at equilibrium, the free energy change of the reaction is zero. Understanding this concept is key to solving problems related to equilibrium and reaction spontaneity.

Equilibrium Constant

The equilibrium constant (K) is a measure of the position of equilibrium in a chemical reaction. It is defined for a reaction in the form:
\[ aA + bB \rightleftharpoons cC + dD \]
The equilibrium constant expression is:
\[ K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]
Square brackets denote concentrations of the reactants and products. For the given reaction:
\[ \text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \]
The equilibrium constant expression is:
\[ K = \frac{[\text{H}_2][\text{O}_2]^\frac{1}{2}}{[\text{H}_2\text{O}]} \]
When K equals 1, it indicates that the concentrations of products and reactants are in a balanced state. This occurs at a specific temperature where the Gibbs free energy change is zero.

Decomposition Reaction

A decomposition reaction involves breaking down a compound into simpler substances. In the context of the given exercise, steam undergoes a decomposition reaction:
\[ \text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \]
The standard enthalpy change (\triangle H^\theta) for this reaction is 200 kJ/mol, and the standard entropy change (\triangle S^\theta) is 40 J/mol·K. To find the temperature at which the equilibrium constant is unity, we use the relationship between Gibbs free energy, enthalpy, and entropy:
\[ \triangle G^\theta = \triangle H^\theta - T \triangle S^\theta \]
When \triangle G^\theta is zero:
\[ 0 = 200 \text{ kJ/mol} - T \times 40 \text{ J/mol·K} \]
Converting 40 J/mol·K to kJ/mol·K (as 1 kJ = 1000 J):
\[ 0 = 200 \text{ kJ/mol} - T \times 0.04 \text{ kJ/mol·K} \]
Solving for T:
\[ T = \frac{200 \text{ kJ/mol}}{0.04 \text{ kJ/mol·K}} = 5000 \text{ K} \]
Hence, the temperature at which the equilibrium constant for the decomposition reaction is unity is 5000 K. This detailed understanding helps in predicting the conditions necessary for specific chemical reactions.