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The equilibrium constant for a reaction is \(10 .\) What will be the value of \(\Delta G^{\circ}\) at \(300 \mathrm{~K} ?\) (a) \(-5.74 \mathrm{~kJ}\) (b) \(-574 \mathrm{~kJ}\) (c) \(+11.48 \mathrm{~kJ}\) (d) \(+5.74 \mathrm{~kJ}\)

Short Answer

Expert verified
-5.74 kJ/mol

Step by step solution

01

Writing the formula to calculate \(\Delta G^{\circ}\)

Determine the Gibbs free energy change \(\Delta G^{\circ}\) for the reaction at equilibrium by using the formula: \[\Delta G^{\circ} = -RT \ln K\] where \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(K\) is the equilibrium constant.
02

Plug in the values

Use the given values to calculate \(\Delta G^{\circ}\). Here, \(K = 10\), \(T = 300\ K\), and the universal gas constant \(R = 8.314\ J/(mol\cdot K)\). Convert \(R\) to kJ by dividing by 1000 if you wish to get your answer in kJ. The equation becomes: \[\Delta G^{\circ} = -\left( \frac{8.314\ J}{mol\cdot K} \right)\left(300\ K\right)\ln(10)\]
03

Calculate \(\Delta G^{\circ}\)

Perform the calculation. Keeping in mind \(\ln(10) \approx 2.303\), \[\Delta G^{\circ} = -\left(\frac{8.314}{1000}\frac{kJ}{mol\cdot K}\right)\left(300\ K\right)(2.303) = -\left(\frac{8.314}{1000}\right)(300)(2.303)\approx -5.74\ kJ/mol\]

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Key Concepts

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

Understanding Equilibrium Constants
When studying chemical reactions, the notion of equilibrium is central. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of products and reactants over time. The equilibrium constant (K) represents the ratio of product concentrations to reactant concentrations at this state, each raised to the power of their respective coefficients in the balanced equation.

The magnitude of K provides insight into the position of equilibrium. A large K (much greater than 1) implies a reaction favoring products; conversely, a small K (much less than 1) suggests that reactants are favored. Understanding this concept is crucial for predicting the extent to which a reaction will proceed under specific conditions.

Calculating K from Gibbs Free Energy

In the context of the exercise, knowing K allows for the determination of the Gibbs free energy change, Delta G°. The relationship between these values is governed by the equation Delta G° = -RT ln(K) where R is the universal gas constant, T is the temperature in Kelvin, and ln(K) is the natural logarithm of the equilibrium constant. Grasping the correlation between Delta G° and K is imperative to predict the spontaneity and directionality of a reaction.
Exploring Chemical Thermodynamics
Chemical thermodynamics involves the study of energy changes accompanying chemical reactions and physical processes. It paves the way for predicting reactant and product behavior under varying conditions of temperature and pressure. One central concept is Gibbs free energy (Delta G°), which represents the maximum amount of work that can be performed by a system at constant temperature and pressure.

Significance of Gibbs Free Energy

Delta G° is a critical indicator of reaction spontaneity; negative values suggest a reaction is spontaneous, while positive values indicate non-spontaneity. When Delta G° equals zero, the system is at equilibrium, and no net work can be done. By calculating Delta G°, you unearth whether a reaction can occur without external energy input.

The provided exercise exemplifies a common thermodynamics problem—calculating Delta G° from a known equilibrium constant to ascertain reaction spontaneity. Comprehending these thermodynamic principles is instrumental in various fields, from developing new materials to understanding biological systems.
Determining Reaction Spontaneity
A spontaneous reaction is one that occurs on its own, without external intervention. However, spontaneity does not equate to speed; rather, it refers to the inherent tendency of a process to occur under a given set of conditions. To determine this tendency, chemists assess the Gibbs free energy change (Delta G°).

A key takeaway is that spontaneity is direction-dependent. For a given reaction occurring at a particular temperature and pressure, if Delta G° is negative, the forward reaction is spontaneous. If Delta G° is positive, the reverse reaction is spontaneous.

Understanding 'Spontaneity' in Thermodynamic Terms

What does this mean in practical terms? A negative Delta G° suggests that a reaction could theoretically initiate and proceed without any external energy—this is often associated with liberating energy (exergonic reaction). On the other hand, a positive Delta G° would require an input of energy to drive the reaction (endergonic reaction).

When approaching problems such as the one in our exercise, a solid understanding of these energy concepts is invaluable. It allows you to predict if a reaction will occur as written (forward direction) or if the system will favor the reactants (reverse direction), straight from the numeric value of Delta G°. The clear interpretation of Delta G° in conjunction with the equilibrium constant provides a full picture of the reaction's characteristics and potential to proceed.

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Most popular questions from this chapter

Match the column I with column II and mark the appropriate choice. $$ \begin{array}{|l|l|l|l|} \hline \text { (A) } & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(g)} & \text { (i) } & \Delta H=\Delta U-2 R T \\ \hline \text { (B) } & \mathrm{PCl}_{5(g)} \rightarrow \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(g)} & \text { (ii) } & \Delta H=\Delta U+3 R T \\ \hline \text { (C) } & \mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{NH}_{3(g)} & \text { (iii) } & \Delta H=\Delta U \\ \hline \text { (D) } & \begin{array}{l} 2 \mathrm{~N}_{2} \mathrm{O}_{5(g)} \rightarrow 4 \mathrm{NO}_{2(g)} \\ +\mathrm{O}_{2(g)} \end{array} & \text { (iv) } & \Delta H=\Delta U+R T \\ \hline \end{array} $$ (a) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (i), (C) \(\rightarrow\) (ii), (D) \(\rightarrow\) (iv) (b) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (iv), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (ii) (c) (A) \(\rightarrow\) (ii), (B) \(\rightarrow\) (i), (C) \(\rightarrow\) (iv), (D) \(\rightarrow\) (iii) (d) (A) \(\rightarrow\) (iv), (B) \(\rightarrow\) (ii), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (iii)

For reversible reaction : \(X_{(g)}+3 Y_{(g)} \rightleftharpoons 2 Z_{(g)} ;\) \(\Delta H=-40 \mathrm{~kJ}\) Standard entropies of \(X, Y\) and \(Z\) are 60,40 and \(50 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) respectively. The temperature at which the above reaction is in equilibrium is (a) \(273 \mathrm{~K}\) (b) \(600 \mathrm{~K}\) (c) \(500 \mathrm{~K}\) (d) \(400 \mathrm{~K}\)

Read the following statements regarding spontaneity of a process and mark the appropriate choice. (i) When enthalpy factor is absent then randomness factor decides spontaneity of a process. (ii) When randomness factor is absent then enthalpy factor decides spontaneity of a process. (iii) When both the factors take place simultaneously, the magnitude of both of factors decide spontaneity of a process. (a) Statements (i) and (ii) are correct and (iii) is incorrect. (b) Statement (iii) is correct, (i) and (ii) are incorrect. (c) Statements (i), (ii) and (iii) are correct. (d) Statements (i), (ii) and (iii) are incorrect.

When the system does not exchange heat with the surroundings, the process is (a) isothermal (b) adiabatic (c) thermal (d) isochoric.

What will be \(\Delta H\) for the reaction, \(\mathrm{CH}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{C}+2 \mathrm{H}+2 \mathrm{Cl} ?\) \(\left(B \cdot E\right.\). of \(\mathrm{C}-\mathrm{H}\) and \(\mathrm{C}-\mathrm{Cl}\) bonds are \(416 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(325 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively) (a) \(832 \mathrm{~kJ}\) (b) \(1482 \mathrm{~kJ}\) (c) \(650 \mathrm{~kJ}\) (d) \(1855 \mathrm{~kJ}\)

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