(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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EBK GENERAL CHEMISTRY
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- Use the data in Appendix J to calculate rG andKPat 25 C for the reaction 2HBr(g)+Cl2(g)2HCl(g)+Br2() Comment on the connection between the sign of rG and the magnitude ofKP.arrow_forwardFor each process, predict whether entropy increases or decreases, and explain how you arrived at your prediction. 2 CO2(g) → 2 CO(g) + O2(g) NaCl(s) → NaCl(aq) MgCO3(s) → MgO(s) + CO2(g)arrow_forwardWhat is the sign of the standard Gibbs free-energy change at low temperatures and at high temperatures for the explosive decomposition of TNT? Use your knowledge of TNT and the chemical equation, particularly the phases, to answer this question. (Thermodynamic data for TNT are not in Appendix G.) 2C7H5N3O6(s) 3N2(g) + 5H2O() + 7C(s) + 7CO(g)arrow_forward
- Elemental 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_forwardThe 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_forward
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