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Chapter 7: Problem 32

find \(\Delta \mathrm{G}^{\circ}\) for the reaction given below? \(\frac{1}{2} \mathrm{~A}+\frac{3}{2} \mathrm{~B} \rightleftharpoons \mathrm{C}_{\mathrm{B}}\) \(\mathrm{K}_{\mathrm{eq}}=826 \mathrm{~atm}^{-1}\) at \(298 \mathrm{~K}\) (a) \(-8.32 \mathrm{KJ}\) (b) \(8.32 \mathrm{KJ}\) (c) \(16.64 \mathrm{KJ}\) (d) \(-16.64 \mathrm{KJ}\)

Short Answer

Expert verified
The value of \( \Delta \mathrm{G}^{\circ} \) is approximately \(-16.64 \, \mathrm{kJ/mol}\), so the correct answer is (d) \(-16.64 \, \mathrm{kJ}\).

Step by step solution

01

Understand the Relationship

To find the standard Gibbs free energy change, \( \Delta \mathrm{G}^{\circ} \), for a reaction, we use the formula \( \Delta \mathrm{G}^{\circ} = -RT \ln K_{eq} \), where \( R \) is the gas constant \( (8.314 \, \mathrm{J/(mol \cdot K)}) \), \( T \) is the temperature in Kelvin, and \( K_{eq} \) is the equilibrium constant.
02

Convert Units

Before substituting into the equation, note that the gas constant \( R \) is given in \( \mathrm{J/(mol \cdot K)} \), so \( \Delta \mathrm{G}^{\circ} \) will also be calculated in \( \mathrm{J} \) initially. We will convert it to \( \mathrm{kJ} \) by dividing by 1000.
03

Substitute Known Values

Substitute \( R = 8.314 \, \mathrm{J/(mol \cdot K)} \), \( T = 298 \, \mathrm{K} \), and \( K_{eq} = 826 \, \mathrm{atm}^{-1} \) into the formula: \[ \Delta \mathrm{G}^{\circ} = -8.314 \, \mathrm{J/(mol \cdot K)} \times 298 \, \mathrm{K} \times \ln(826) \].
04

Compute the Logarithm

Calculate \( \ln(826) \). Using a calculator, \( \ln(826) \approx 6.716 \).
05

Calculate the Gibbs Free Energy

Substitute \( \ln(826) \approx 6.716 \) back into the equation: \[ \Delta \mathrm{G}^{\circ} = -8.314 \, \mathrm{J/(mol \cdot K)} \times 298 \, \mathrm{K} \times 6.716 \].
06

Final Calculation

Perform the multiplication: \( \Delta \mathrm{G}^{\circ} = -8.314 \times 298 \times 6.716 \approx -16644.24 \, \mathrm{J/mol} \). Convert this into \( \mathrm{kJ/mol} \) by dividing by 1000: \( \Delta \mathrm{G}^{\circ} \approx -16.64 \, \mathrm{kJ/mol} \).
07

Choose the Correct Answer

The calculated standard Gibbs free energy change \( \Delta \mathrm{G}^{\circ} \approx -16.64 \, \mathrm{kJ/mol} \) corresponds to choice (d) \(-16.64 \, \mathrm{kJ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant and its Role
The equilibrium constant, denoted as \( K_{eq} \), acts as a bridge between thermodynamics and chemical reactions to determine the extent of a reaction at equilibrium. If you have ever wondered why some reactions seem to stop progressing, it's because they've reached a balance point where the rate of the forward reaction equals the rate of the backward reaction.
  • For a chemical reaction at equilibrium, \( K_{eq} \) quantifies this balance. It's expressed as the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced chemical equation.
  • A large \( K_{eq} \) (like 826 atm-1 in our problem) means the products are favored at equilibrium, implying the reaction makes a significant amount of products.
  • The unit of \( K_{eq} \) changes depending on the nature of the reaction - in this case, atm-1, suitable for gaseous equilibrium reactions.
Understanding \( K_{eq} \) is crucial because it helps deduce how much energy is involved when reaching equilibrium, which connects directly to the concept of Gibbs free energy.
Thermodynamics and Gibbs Free Energy
Thermodynamics is the study of energy transformations and it is critical in understanding chemical reactions. One of its key concepts is Gibbs free energy, \( \Delta G \), which measures the spontaneity of a reaction.
  • When \( \Delta G \) is negative, the process is spontaneous. A positive \( \Delta G \) indicates non-spontaneity, and zero denotes equilibrium.
  • The standard Gibbs free energy change, \( \Delta G^{\circ} \), is calculated using the formula: \( \Delta G^{\circ} = -RT \ln K_{eq} \). Here, \( R \) is the universal gas constant, \( T \) is temperature in Kelvin, and \( \ln K_{eq} \) signifies the natural logarithm of the equilibrium constant.
  • This relationship highlights that the larger the \( K_{eq} \), the more negative \( \Delta G^{\circ} \) becomes, reinforcing that a reaction with significant product production is highly spontaneous.
This is why thermodynamics, specifically the concept of Gibbs free energy, is so integral to predicting reaction behavior and conditions under which reactions are favorable.
Logarithms in Chemistry
Logarithms may seem purely mathematical, but they play a vital role in chemistry, particularly when dealing with large numbers, such as equilibrium constants. They simplify complex calculations.
  • The logarithmic function, particularly the natural logarithm \( \ln \), is used because it allows us to linearize exponential growth (or decay) often encountered in chemical equilibria.
  • In the context of calculating \( \Delta G^{\circ} \), knowing \( \ln K_{eq} \) turns the exponential dependence into a linear one, making it easier to manage. For example, \( \ln(826) \approx 6.716 \).
  • This transformation is essential for determining how sensitive equilibrium states are to changes in energy or other conditions, facilitating more straightforward predictions of chemical behavior.
Understanding logarithms helps in unraveling the effects of equilibrium constant values on reaction dynamics, making it an indispensable tool in chemical thermodynamics.

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Most popular questions from this chapter

A saturated solution of non-radioactive sugar was taken and a little radioactive sugar was added to it. A small amount of it gets dissolved in solution and an equal amount of sugar was precipitated. This proves (a) the equilibrium has been established in the solution (b) radioactive sugar can displace non-radioactive sugar from its solution. (c) Equilibrium is dynamic in nature (d) none of the above

For the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) the forward reaction at constant temperature is favoured by 1\. introducing an inert gas at constant volume 2\. introducing chlorine gas at constant volume 3\. introducing an inert gas at constant pressure 4\. increasing the volume of the container 5\. introducing \(\mathrm{PCl}_{5}\) at constant volume (a) \(1,2,3\) (b) 4,5 (c) \(2,3,5\) (d) \(3,4,5\)

If for the reaction given below \(2 \mathrm{PQ} \rightleftharpoons \mathrm{P}_{2}+\mathrm{Q}_{2} \mathrm{~K}_{1}=2.5 \times 10^{5}\) \(\mathrm{PQ}+\frac{1}{2} \mathrm{R}_{2} \rightleftharpoons \mathrm{PQR} \mathrm{K}_{2}=5 \times 10^{-3}\) find \(K_{3}\) for the reaction \(\frac{1}{2} P_{2}+\frac{1}{2} Q_{2}+\frac{1}{2} R_{2} \rightleftharpoons P Q R\) (a) \(2.5 \times 10^{-3}\) (b) \(2.5 \times 10^{3}\) (c) \(1 \times 10^{-3}\) (d) \(5 \times 10^{-3}\)

At \(550 \mathrm{~K}\), the \(\mathrm{K}\) for the following reaction is \(10^{4} \mathrm{~mol}^{-1}\) lit \(\mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons{ }^{\mathrm{c}}-\mathrm{Z}(\mathrm{g})\) At equilibrium, it was observed that \([\mathrm{X}]=1 / 2[\mathrm{Y}]=1 / 2[\mathrm{Z}]\) What is the value of \([\mathrm{Z}]\left(\right.\) in \(\left.\mathrm{mol} \mathrm{L}^{-1}\right)\) at equilibrium? (a) \(2 \times 10^{-4}\) (b) \(10^{-4}\) (c) \(2 \times 10^{4}\) (d) \(10^{4}\)

For a gaseous reaction \(2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\), the partial pressures of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) at equilibrium are \(0.5\), \(0.8 .0 .7\) and \(1.2\) atm. The value of \(K_{p}\) for this reaction is (a) \(2.4 \mathrm{~atm}\) (b) \(6.2 \mathrm{~atm}^{-2}\) (c) \(4.2 \mathrm{~atm}^{-1}\) (d) \(8.4 \mathrm{~atm}^{-3}\)
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