Consider the formation of NO(g) from its elements.
N2(g) + O2(g) ⇄ 2 NO(g)
- (a) Calculate Kp at 25 °C. Is the reaction product-favored at equilibrium at this temperature?
- (b) Assuming ΔrH° and ΔrS° are nearly constant with temperature, calculate ΔrG° at 700 °C. Estimate Kp from the new value of ΔrG° at 700 °C. Is the reaction product-favored at equilibrium at 700 °C?
- (c) Using Kp at 700 °C, calculate the equilibrium partial pressures of the three gases if you mix 1.00 bar each of N2 and O2.
(a)
Interpretation:
The equilibrium constant for formation of
Concept introduction:
The Gibbs free energy or the free energy change is a thermodynamic quantity represented by
The rearranged expression is,
It is also related to entropy and entropy by the following expression,
Here,
Answer to Problem 90SCQ
The value of
The reaction is reactant-favoured at equilibrium.
Explanation of Solution
The value of
Given:
The Appendix L referred for values of standard free energy values.
The given reaction is,
The
The
The
Substituting the respective values,
The value of
The rearranged expression is,
Substituting the respective values of
(b)
Interpretation:
The
Concept introduction:
The Gibbs free energy or the free energy change is a thermodynamic quantity represented by
The rearranged expression is,
It is also related to entropy and entropy by the following expression,
Here,
Answer to Problem 90SCQ
The value of
The value of
Explanation of Solution
The value of
Given:
The Appendix L referred for values of standard entropies and enthalpies.
Substitute the values,
Substitute the values,
Now,
Substitute the value of
The value of
The rearranged expression is,
Substitute the value of
(c)
Interpretation:
The equilibrium partial pressures of the given three gases should be calculated if they were mixed
Concept introduction:
The Gibbs free energy or the free energy change is a thermodynamic quantity represented by
The rearranged expression is,
It is also related to entropy and entropy by the following expression,
Here,
Answer to Problem 90SCQ
The equilibrium partial pressure of
The equilibrium partial pressure of
The equilibrium partial pressure of
Explanation of Solution
The equilibrium partial pressure of
Given:
The initial pressures of
Now,
The equilibrium constant is related to the equilibrium partial pressure by the expression,
Substitute the values,
Thus, the value of
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Chapter 18 Solutions
Chemistry & Chemical Reactivity
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- Elemental boron, in the form of thin fibers, can be made by reducing a boron halide with H2. BCl3(g) + 32 H2(g) B(s) + 3 HCl(g) Calculate rH, rS, and rG at 25 C for this reaction. Is the reaction predicted to be product-favored at equilibrium at 25 C? If so, is it enthalpy- or entropy-driven? [S for B(s) is 5.86 J/K mol.]arrow_forwardBenzene can be prepared from acetylene. 3C2H2(g)C6H6(g). Determine the equilibrium constant at 25 C and at 850 C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?arrow_forwardWithout doing any calculations, predict the sign of rS for the following reaction: Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2(g) (a) rS 0 (b) rS = 0 (c) rS 0arrow_forward
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- Which contains greater entropy, a quantity of frozen benzene or the same quantity of liquid benzene at the same temperature? Explain in terms of the dispersal of energy in the substance.arrow_forwardAt room temperature, the entropy of the halogens increases from I2 to Br2 to Cl2. Explain.arrow_forward
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