At 700 K. the reaction 2502la) - Ozlal = 2503la) has the equilibrium constant Ke -4.3 - 106, and the following concentrations are present. [SO,1 - 0.010 M. [SO3l- 10. M. [02)-0.010 M. Which of the following is true based on the above? Multiple Choice Qc Ke, the reaction proceeds from right t10 left o reach equibrium Qc "Ko the reection is curenty at equilbrium Q >K, the reection proceeds from left to nght to reach equibrtum Q cke the reaction proceeds from let to right to reach equlibrium

At 700 K. the reaction 2502la) - Ozlal = 2503la) has the equilibrium constant Ke -4.3 - 106, and the following concentrations are present. [SO,1 - 0.010 M. [SO3l- 10. M. [02)-0.010 M. Which of the following is true based on the above? Multiple Choice Qc Ke, the reaction proceeds from right t10 left o reach equibrium Qc "Ko the reection is curenty at equilbrium Q >K, the reection proceeds from left to nght to reach equibrtum Q cke the reaction proceeds from let to right to reach equlibrium

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 51P

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A common type of reaction we will study is that having a very small K value (K 1). Solving for equilibrium concentrations in an equilibrium problem usually requires many mathematical operations to be perfomed. However, the math involved when solving equilibrium problems for reactions having small K values (K 1) is simplified. What assumption is made when solving the equilibrium concentrations for reactions with small K values? Whenever assumptions are made, they must be checked for validity. In general, the 5% rule is used to check the validity of assuming x (or 2 x, 3x, and so on) is very small compared to some number. When x (or 2 x. 3x. and so on) is less than 5% of the number the assumption was made against, then the assumption is said to be valid. If the 5% rule fails, what do you do to solve for the equilibrium concentrations?
At 2000 K, experiments show that the equilibrium constant for the formation of water is 1.6 1010. 2H2(g) + O2(g) 2H2O(g) Calculate the equilibrium constant at the same temperature for H2(g)+12O2(g) H2O(g)
Consider the decomposition equilibrium for dinitrogen pentoxide: 2N2O5(g)4NO2(g)+O2(g) At a certain temperature and a total pressure of 1.00 atm, the N2O5 is 0.50% decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of 10.0. will the mole percent of N2O5 decomposed at equilibrium be greater than, less than. or equal to 0.50%? Explain your answer. b. Calculate the mole percent of N2O5 that will be decomposed at equilibrium if the volume is increased by a factor of 10.0.
7-64 As we shall see in Chapter 20, there are two forms of glucose, designated alpha and betawhich are in equilibrium in aqueous solution. The equilibrium constant for the reaction is 1.5 at 30°C. (a) If you begin with a fresh 1.0 M solution of D-glucose in water, what will be its concentration when equilibrium is reached? (b) Calculate the percentage of glucose and of glucose present at equilibrium in aqueous solution at 30°C.
12.42 The following reaction is in equilibrium in lake water: HCO,_(aq) + H+(aq)«=*H,CO,(aq) Predict the change in the reaction quotient, Q, for each disturbance below and use that prediction to explain how the equilibrium is shifted by the stress. NaHCOj is added to the lake. H,CO}is added. NaOH is added.
Nitrogen, hydrogen, and ammonia are in equilibrium in a 1000-L reactor at 550 K. The concentration of N2 is 0.00485 M, H2 is 0.022 M, and NH3 is 0.0016 M. The volume of the container is halved, to 500 L. (a) Calculate the equilibrium constant. (b) Define y as the change in the concentration of nitrogen in the 500-L container. Is y positive or negative? (c) Write the iCe table and express the equilibrium constant in a polynomial in terms of y.
Consider the decomposition of ammonium hydrogen sulfide: NH4HS(s)NH3(g)+H2S(g) In a sealed flask at 25C are 10.0 g of NH4HS, ammonia with a partial pressure of 0.692 atm, and H2S with a partial pressure of 0.0532 atm. When equilibrium is established, it is found that the partial pressure of ammonia has increased by 12.4%. Calculate K for the decomposition of NH4HS at 25C.
7-29 The following reaction was allowed to reach equilibrium at 25°C. Under each component is its equilibrium concentration. Calculate the equilibrium constant, K, for this reaction.
Many sugars undergo a process called mutarotation, in which the sugar molecules interconvert between two isomeric forms, finally reaching an equilibrium between them. This is true for the simple sugar glucose, C6H12O6, which exists in solution in isomeric forms called alpha-glucose and beta-glucose. If a solution of glucose at a certain temperature is analyzed, and it is found that the concentration of alpha-glucose is twice the concentration of beta-glucose, what is the value of K for the interconversion reaction?
Because carbonic acid undergoes a second ionization, the student in Exercise 12.39 is concerned that the hydrogen ion concentration she calculated is not correct. She looks up the equilibrium constant for the reaction HCO,-(aq) «=* H+(aq) + COf'(aq) Upon finding that the equilibrium constant for this reaction is 4.8 X 10“H, she decides that her answer in Exercise 12.39 is correct. Explain her reasoning. A student is simulating the carbonic acid—hydrogen carbonate equilibrium in a lake: H,CO,(aq) 5=6 H+(aq) + HCO,'(aq) K = 4.4 X 10'7She starts with 0.1000 A1 carbonic acid. W hat are the concentrations of all species at equilibrium?
For the system SO3(g)SO2(g)+12 O2(g)at 1000 K, K=0.45. Sulfur trioxide, originally at 1.00 atm pressure, partially dissociates to SO2 and O2 at 1000 K. What is its partial pressure at equilibrium?
Consider the system A(g)+2B(g)+C(g)2D(g)at 25C. At zero time, only A, B, and C are present. The reaction reaches equilibrium 10 min after the reaction is initiated. Partial pressures of A, B, and D are written as PA, PB, and PD. Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required). (a) PD at 11 min PD at 12 min. (b) PA at 5 min PA at 7 min. (c) K for the forward reaction K for the reverse reaction. (d) At equilibrium, K Q. (e) After the system is at equilibrium, more of gas B is added. After the system returns to equilibrium, K before the addition of B K after the addition of B. (f) The same reaction is initiated, this time with a catalyst K for the system without a catalyst the system with a catalyst. (g) K for the formation of one mole of D K for the formation of two moles of D. (h) The temperature of the system is increased to 35C. PB at equilibrium at 25C PB at equilibrium at 35C. (i) Ten more grams of C are added to the system. PB before the addition of C PB after the addition of C.