Question

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at \(600 . \mathrm{K}\) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at 600. \(\mathrm{K}\)

Short answer

The standard Gibbs free energy change, \(\Delta G^\circ\), for the reaction \(\mathrm{H_2O(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O_2(g)}\) at 600. K is approximately \(-67162.3 \ J \cdot mol^{-1}\).

Step-by-step solution

Obtain the desired reaction through manipulations of the given reactions

First, we will subtract half of the first reaction from the second reaction to obtain the desired reaction: \[2\mathrm{H_2(g) + O_2(g) \rightleftharpoons 2H_2O(g)}\] \[-\frac{1}{2}(\mathrm{H_2(g) + O_2(g) \rightleftharpoons H_2O_2(g)})\] Adding both equations, we get: \[\mathrm{H_2O(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O_2(g)}\]

Calculate the equilibrium constant for the desired reaction

We can find the equilibrium constant, \(K\), for the desired reaction by manipulating the given equilibrium constants \(K_1\) and \(K_2\). Since we have subtracted half of the first reaction from the second reaction: \[K = \frac{K_2}{K_1^{\frac{1}{2}}}\] Substitute the given values for \(K_1\) and \(K_2\): \[K = \frac{1.8 \times 10^{37}}{(2.3 \times 10^{6})^{\frac{1}{2}}}\] Calculate the value of K: \[K \approx 2.451 \times 10^{30}\]

Calculate the standard Gibbs free energy change

Now, we can use the formula for calculating the standard Gibbs free energy change (\(\Delta G^\circ\)) from the equilibrium constant: \[\Delta G^\circ = -RT\ln{K}\] Where \(R\) is the gas constant (8.314 JK^{-1}mol^{-1}), \(T\) is the temperature in Kelvin (600 K), and \(K\) is the equilibrium constant we found in the previous step. Substitute the values: \[\Delta G^\circ = -(8.314)(600) \ln{(2.451 \times 10^{30})}\] Perform the calculation: \[\Delta G^\circ \approx -67162.3 \ J \cdot mol^{-1}\] So, the standard Gibbs free energy change, \(\Delta G^\circ\), for the reaction \(\mathrm{H_2O(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O_2(g)}\) at 600. K is approximately \(-67162.3 \ J \cdot mol^{-1}\).

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \(K\), is a crucial concept in chemical reactions that indicates the ratio of the concentrations of products to reactants at equilibrium. For a reaction at a specific temperature, \(K\) remains constant and helps predict the direction in which a reaction will proceed.In this exercise, different reactions provided have their own equilibrium constants:

• The conversion of hydrogen and oxygen to hydrogen peroxide has \(K_1 = 2.3 \times 10^6\).
• The conversion to water has a much larger \(K_2 = 1.8 \times 10^{37}\).

Manipulating these reactions to find the equilibrium constant for the desired reaction involves the use of these constants. When reactions are combined, their equilibrium constants can be adjusted mathematically to reflect the changes in reaction stoichiometry. Here, the reaction was manipulated by subtracting half of one reaction from another. Thus, the new equilibrium constant is derived using this relationship:\[K = \frac{K_2}{K_1^{0.5}}\]

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. It describes how energy moves and transforms in chemical reactions, playing a crucial role in these transformations.One essential aspect is the Gibbs free energy change (\(\Delta G\)) which indicates whether a process occurs spontaneously. A negative \(\Delta G\) suggests a spontaneous reaction under constant pressure and temperature.In this problem, the standard Gibbs free energy change \(\Delta G^\circ\) is calculated from the equilibrium constant:\[\Delta G^\circ = -RT \ln K\]

• \(R\) is the universal gas constant \(8.314 \text{ J K}^{-1}\text{mol}^{-1}\).
• \(T\) is the temperature in Kelvin \(600 \text{ K}\).
• \(K\) is the equilibrium constant \(2.451 \times 10^{30}\).

The negative \(\Delta G^\circ\) calculated confirms that the reaction can proceed spontaneously at standard conditions.

Reaction Manipulation

Reaction manipulation involves adjusting chemical equations to derive new reactions. This is done through additions and subtractions of given reactions, and is a common method used to find desired chemical outcomes or reactions.In this exercise, to find the equilibrium constant \(K\) of the desired reaction, known reactions were manipulated:

• Subtraction of half of one reaction from another.
• This manipulation changes the stoichiometry, hence impacting the equilibrium constant.
• Mathematically, this required division by the square root of one equilibrium constant due to halving, as shown in the formula \(K = \frac{K_2}{K_1^{0.5}}\).

Such manipulation is key in thermodynamics to determine thermodynamic quantities like \(\Delta G^\circ\), making it a handy tool for chemists to evaluate different chemical processes.