Question

Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each reaction: (a) \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) ; K=5.62 \times 10^{35}\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{3}(g) ; K=4.46 \times 10^{-15}\)

Short answer

The term 298 translates to Kelvin. At Twith Δtemperature tem <|vq_7777|> However, the explanation must have the correct sign for better explanation opportunities, leading to the Kelvin conversion failure for final Gibbs energy calculation, leading to large-scale predictions; have lower magnitude/functions per math equation

Step-by-step solution

- Understand the Gibbs Free Energy Equation

The Gibbs Free Energy change for a reaction at standard conditions can be calculated using the equation: Equation: $$\text{

Key concepts

Gibbs Free Energy

Gibbs Free Energy, often referred to as \( \Delta G \), is a thermodynamic function that helps predict whether a reaction will proceed spontaneously under constant pressure and temperature. The fundamental equation to calculate Gibbs Free Energy change is: \( \Delta G = \Delta H - T \Delta S \). However, at standard conditions, we typically use a more specific form to relate \( \Delta G^{\circ} \) to the equilibrium constant (K). This form of the equation is expressed as: \( \Delta G^{\circ} = -RT \ln(K) \). Here, \( R \) is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvins, and K is the equilibrium constant.
Understanding \( \Delta G \) is crucial because it tells us:

• If \( \Delta G < 0 \), the reaction is spontaneous and will proceed.
• If \( \Delta G = 0 \), the system is at equilibrium.
• If \( \Delta G > 0 \), the reaction is non-spontaneous.

Calculating \( \Delta G^{\circ} \) using the equilibrium constant helps us understand the spontaneity of a reaction under standard conditions.

Equilibrium Constant

The equilibrium constant, represented as K, is a vital part of chemical reaction dynamics. It indicates the ratio of the concentrations of products to reactants at equilibrium for a given reaction.
The equilibrium constant can be expressed either in terms of concentration ( \( K_{c} \)) or partial pressure ( \( K_{p} \)), depending on the state of the reactants and products. The key point about K is:

• If \( K > 1 \), products are favored at equilibrium.
• If \( K < 1 \), reactants are favored at equilibrium.

In the context of calculating \( \Delta G^{\circ} \), the value of K directly influences the sign and magnitude of \( \Delta G^{\circ} \). For example, a very large K (as seen in part (a) of the exercise with \( K = 5.62 \times 10^{35} \)) suggests that the reaction heavily favors the formation of products, leading to a highly negative \( \Delta G^{\circ} \), which indicates a spontaneous process. Conversely, a very small K (as in part (b) with \( K = 4.46 \times 10^{-15} \)) suggests a reaction heavily biased towards the reactants, resulting in a positive \( \Delta G^{\circ} \), indicating non-spontaneity.

Standard Conditions

Standard conditions, denoted by the superscript \( ^{\circ} \), refer to a set of specified criteria that simplify the comparison of different reactions. These parameters include:

• A temperature of 298 K (25°C).
• A pressure of 1 atm for gases.
• A concentration of 1 M for solutions.

In thermodynamic calculations, including those for \( \Delta G^{\circ} \), standard conditions allow consistency and comparability across different reactions.
When we calculate \( \Delta G^{\circ} \) at standard conditions using the equation \( \Delta G^{\circ} = -RT \ln(K) \), we assume that the temperature, pressure, and concentrations are all aligned with these standard benchmark values. This allows us to predict the spontaneity of reactions under a common set of conditions, making the thermodynamic parameters more universally applicable and understandable.