Concept explainers
(a)
Interpretation:
The equilibrium constant at
Concept Introduction:
The equilibrium constant of a reaction can be calculated using the given expression,
(a)
Answer to Problem 5G.21E
The equilibrium constant of the given reaction is
Explanation of Solution
Given reaction is
The combustion of hydrogen:
Temperature of the reaction is
The
The
The
Here,
Now, the equilibrium constant of the reaction is calculated as
To calculate
Therefore, the
(b)
Interpretation:
The equilibrium constant at
Concept introduction:
Refer to part (a).
(b)
Answer to Problem 5G.21E
The equilibrium constant of the given reaction is
Explanation of Solution
Given reaction is
The oxidation of carbon monoxide:
Temperature of the reaction is
The
The
The
Here,
Now, the equilibrium constant of the reaction is calculated as
To calculate
Therefore, the
(c)
Interpretation:
The equilibrium constant at
Concept introduction:
Refer to part (a).
(c)
Answer to Problem 5G.21E
The equilibrium constant of the given reaction is
Explanation of Solution
Given reaction is
The decomposition of limestone:
Temperature of the reaction is
The
The
The
Here,
Now, the equilibrium constant of the reaction is calculated as
To calculate
Therefore, the
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Chapter 5 Solutions
Chemical Principles: The Quest for Insight
- At a certain temperature, K=0.29 for the decomposition of two moles of iodine trichloride, ICl3(s), to chlorine and iodine gases. The partial pressure of chlorine gas at equilibrium is three times that of iodine gas. What are the partial pressures of iodine and chlorine at equilibrium?arrow_forwardDescribe a nonchemical system that is not in equilibrium, and explain why equilibrium has not been achieved.arrow_forwardActually, the carbon in CO2(g) is thermodynamically unstable with respect to the carbon in calcium carbonate(limestone). Verify this by determining the standardGibbs free energy change for the reaction of lime,CaO(s), with CO2(g) to make CaCO3(s).arrow_forward
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- The equilibrium constant for the following reaction is 3.13x103 at 32 °C. N2(g) + O2(g) =2 NO(g) = 3.13×103 at 32 °C Calculate the equilibrium constant for the following reactions at 32 °C. (a) 2 NO(g) = N2(g) + 02(g) K = (b) NO(g)=1/2 N2(g) + 1/2 02(g) K =arrow_forwardPhosphoryl chloride, POCI3(g), is used in the manufacturing of flame retardants. It is manufactured in an equilibrium process in which phosphorus trichloride reacts with nitrogen dioxide to form POCI3(g)and NO(g) according to the following equation: PCI3 (g) + NO2 (g) = POCI; (g) + NO (g) At a certain temperature, the equilibrium concentrations of POCI3 and NO was 0.79 mol/L, and the concentrations of PCI3 and NO2 was 0.90 mol/L. At this temperature the Keg is 4.65. 4.75 moles of NO2 (g) is added to the 1.0 L reaction chamber. What is the concentration of POCI3 (g) when equilibrium is re-established? PCI3 (g) NO2 (g) POCI3 (g) NO(g) + + E' Earrow_forwardConsider the reaction: 2 CO(g) + O2(g)=2 CO₂(g). The reaction is allowed to reach equilibrium in a sealed vessel. According to Le Chatelier's principle, what will happen to the equilibrium, if the volume of the vessel is decreased while the temperature is kept constant? (A) The equilibrium constant will decrease and the reaction will shift to the left. (B) The equilibrium constant will be unchanged, but the reaction will shift to the left. (C) The equilibrium constant will be unchanged, but the reaction will shift to the right. (D) The equilibrium constant will increase and the reaction will shift to the right. (E) The equilibrium concentrations will not be affected.arrow_forward
- Consider the equilibrium4NO2(g) + 6H2 O(g) ⇌ 4NH3(g) + 7O2(g)(a) What is the expression for the equilibrium constant (Kc) of the reaction?(b) How must the concentration of NH3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?(c) If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the volume of the reaction vessel) affect the pressure of NO2?(d) If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change?arrow_forwardConsider the equilibrium system described by the chemical reaction below. If 8.98 × 104 moles of KOH(aq), 8.98 × 104 moles of HCN(aq), and 0.0500 moles of KCN(aq) are present in a 1.00 L aqueous solution at equilibrium at 298 K, what is the value of K of the reaction at this temperature? KCN(aq) + H,O(I) = KOH(aq) + HCN(aq) [8.98 × 104] K || [0.0500] [1.80 x 10-²] 1 [8.06 x 10-7] 1.80 x 10-² RESET 1.61 x 10-5arrow_forwardWrite the mathematical expression for the reaction quotient, Qc, for each of the following reactions: (a) CH4(g)+Cl2(g)⇌CH3Cl(g)+HCl(g)CH4(g)+Cl2(g)⇌CH3Cl(g)+HCl(g) (b) N2(g)+O2(g)⇌2NO(g)N2(g)+O2(g)⇌2NO(g) (c) 2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g) (d) BaSO3(s)⇌BaO(s)+SO2(g)BaSO3(s)⇌BaO(s)+SO2(g) (e) P4(g)+5O2(g)⇌P4O10(s)P4(g)+5O2(g)⇌P4O10(s) (f) Br2(g)⇌2Br(g)Br2(g)⇌2Br(g) (g) CH4(g)+2O2(g)⇌CO2(g)+2H2O(l)CH4(g)+2O2(g)⇌CO2(g)+2H2O(l) (h) CuSO4⋅5H2O(s)⇌CuSO4(s)+5H2O(g)arrow_forward
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