(a)
Interpretation:
To calculate
Concept introduction:
The Gibb’s equation of
With the help of this equation, one can predict the change in
Here, ‘a’ represents the active mass and ‘P’ represents the partial pressure.
The relation between equilibrium constant and
Here,
- R = 8.314 J / mol .K
- T = temperature in Kelvin
(b)
Interpretation:
To calculate
Concept introduction:
The Gibb’s equation of thermodynamic purposed a relation between
With the help of this equation, one can predict the change in
Here, ‘a’ represents the active mass and ‘P’ represents the partial pressure.
The relation between equilibrium constant and
Here,
- R = 8.314 J / mol .K
- T = temperature in Kelvin
(c)
Interpretation:
To calculate
Concept introduction:
The Gibb’s equation of thermodynamic purposed a relation between
With the help of this equation, one can predict the change in
Here, ‘a’ represents the active mass and ‘P’ represents the partial pressure.
The relation between equilibrium constant and
Here,
- R = 8.314 J / mol .K
- T = temperature in Kelvin
(d)
Interpretation:
To calculate
Concept introduction:
The Gibb’s equation of thermodynamic purposed a relation between
With the help of this equation, one can predict the change in
Here, ‘a’ represents the active mass and ‘P’ represents the partial pressure.
The relation between equilibrium constant and
Here,
- R = 8.314 J / mol .K
- T = temperature in Kelvin
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General Chemistry: Principles and Modern Applications - With Solutions Manual and Modified MasteringChemistry Code
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- Use S values to calculate the standard entropy change, rS0, for each of the following processes and comment on the sign of the change. (a) KOH(s) KOH(aq) (b) Na(g) Na(s) (c) Br2() Br2(g) (d) HCl(g) HCl(aq)arrow_forwardConsider the reaction of 1 mol H2(g) at 25C and 1 atm with 1 mol Br2(l) at the same temperature and pressure to produce gaseous HBr at these conditions. If this reaction is run in a controlled way to generate work, what is the maximum useful work that can be obtained? How much entropy is produced in this case?arrow_forwardElemental boron, in the form of thin fibers, can be made by reducing a boron halide with H2. BCl3(g) + 3/2 H2(g) B(s) + 3HCl(g) Calculate H, S, and G at 25 C for this reaction. Is the reaction predicted to be product favored at equilibrium at 25 C? If so, is it enthalpy driven or entropy driven?arrow_forward
- The decomposition of diamond to graphite [C(diamond) C(graphite)] is thermodynamically favored, but occurs slowly at room temperature. a. Use fG values from Appendix L to calculate rG and Keq for the reaction under standard conditions and 298.15 K. b. Use fH and S values from Appendix L to estimate rG and Keq for the reaction at 1000 K. Assume that enthalpy and entropy values are valid at these temperatures. Does heating shift the equilibrium toward the formation of diamond or graphite? c. Why is the formation of diamond favored at high pressures? d. The phase diagram shows that diamond is thermodynamically favored over graphite at 20,000 atmospheres pressure (about 2 GPa) at room temperature. Why is this conversion actually done at much higher temperatures and pressures?arrow_forwardDiscuss the effect of temperature change on the spontaneity of the following reactions at 1 atm: (a) Al2O3(s)+2Fe(s)2Al(s)+Fe2O3(s) H =851.4kJ;S =38.5J/K (b) N2H4(l)N2+2H2(g) H =50.6kJ;S =0.3315kJ/K (c) SO3(g)SO2(g)+12 O2(g) H =98.9kJ;S =0.0939kJ/Karrow_forward
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