(a)
Interpretation:
The value of
Concept Introduction:
The mathematical expression for the standard entropy value at room temperature is:
Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.
The mathematical expression for the standard enthalpy change value at room temperature is:
Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.
Spontaneity depends upon the temperature and also depends upon the sign of free energy change.
The mathematical expression for
If both
When the magnitude of
Therefore, reaction is non- spontaneous at low temperature and spontaneous at high temperature.
(a)
Answer to Problem 3E
Explanation of Solution
The given reaction is:
The value of standard entropy for
The value of standard entropy for
The value of standard entropy for
Put the values, in below formula.
The value of standard enthalpy for
The value of standard enthalpy for
The value of standard enthalpy for
Put the values, in below formula.
The value of standard Gibbs free energy for
The value of standard Gibbs free energy for
The value of standard Gibbs free energy for
Put the values, in below formula.
(b)
Interpretation:
Whether the reaction is spontaneous or not should be determined.
Concept Introduction:
The mathematical expression for the standard entropy value at room temperature is:
Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.
The mathematical expression for the standard enthalpy change value at room temperature is:
Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.
Spontaneity depends upon the temperature and also depends upon the sign of free energy change.
The mathematical expression for
If both
When the magnitude of
Therefore, reaction is non- spontaneous at low temperature and spontaneous at high temperature.
(b)
Answer to Problem 3E
For the given reaction,
Explanation of Solution
The given reaction is:
From part (a):
The negative value of Gibbs free energy represents the given reaction is spontaneous.
(c)
Interpretation:
The temperature at which reaction is spontaneous at standard conditions should be determined by considering
Concept Introduction:
The mathematical expression for the standard entropy value at room temperature is:
Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.
The mathematical expression for the standard enthalpy change value at room temperature is:
Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.
Spontaneity depends upon the temperature and also depends upon the sign of free energy change.
The mathematical expression for
If both
When the magnitude of
Therefore, reaction is non- spontaneous at low temperature and spontaneous at high temperature.
(c)
Answer to Problem 3E
The temperature at which reaction is spontaneous must be greater than
Explanation of Solution
The given reaction is:
From part (a)
Here, sign of
Put the values,
Now,
Let
Since, 1 Kilojoule = 1000 Joule
The process is spontaneous when
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Chapter 18 Solutions
Chemical Principles
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- Using data from Appendix 4, calculate H, G, and K (at 298 K) for the production of ozone from oxygen: 3O2(g)2O3(g) At 30 km above the surface of the earth, the temperature is about 230. K and the partial pressure of oxygen is about 1.0 103 atm. Estimate the partial pressure of ozone in equilibrium with oxygen at 30 km above the earth's surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.arrow_forwardUsing values of fH and S, calculate the standard molar free energy of formation, fG, for each of the following: (a) Ca(OH)2(s) (b) Cl(g) (c) Na2CO3(s) Compare your calculated values of fG with those listed in Appendix L. Which of these formation reactions are predicted to be product-favored at equilibrium at 25 C?arrow_forwardGiven: P4(s)+5O2(g)P4O10(s)G298=2697.0kJ/mol 2H2(g)+O2(g)2H2O(g)G298=457.18kJ/mol 6H2O(g)+P4O10(s)4H3PO4(l)G298=428.66kJ/mol (a) Determine the standard free energy of formation, Gf, for phosphoric acid. (b) How does your calculated result compare to the value in Appendix G? Explain.arrow_forward
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