Write the equilibrium constant expression (K.) for each of the following reactions.а. 302(9) 203(9)03(9)b. Oz0) =030)

Write equilibrium constant expression for each reaction--

Answered: Write the equilibrium constant…

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ChapterU6: Showtime: Reversible Reactions And Chemical Equilibrium

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Problem 2RE

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Write the equilibrium constant expression for each reaction in terms of activities, simplifying where appropriate. a C(s)+O2(g)CO2(g) b P4(s)+5O2(g)P4O10(s) c 2HNO2(g)+3Cl2(g)2NCl3(g)+H2(g)+2O2(g)
1’he reaction in Exercise 12.33 was repeated. This time, the reaction began when only NO was injected into the reaction container. 110.200 mol L_l NO was injected, what were the equilibrium concentrations of all species? The following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) ^2 NO K = 4.1 X 10~4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L"' ofO2, what were the equilibrium concentrations of all species?
Explain how to recognize the conditions under which changes in pressure would affect systems at equilibrium.
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?
The experiment in Exercise 12.33 was redesigned so that the reaction started with 0.15 mol each of N2 and O2 being injected into a 1.0-L container at 2500 K. The equilibrium constant at 2500 K is 3.6 X 10“’. What was the composition of the reaction mixture after equilibrium was attained? The following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) *2 2 NO K = 4.1 X IO-4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L-’ ofO2, what were the equilibrium concentrations of all species?
The following two diagrams represent the composition of an equilibrium mixture for the reaction A2 + B2 2AB at two different temperatures. Based on the diagrams, is the chemical reaction endothermic or exothermic? Explain your answer using Le Chteliers principle. (A atoms are red and B atoms are green in the diagrams.)
For the reactionH2(g)+I2(g)2HI(g), consider two possibilities: (a) you mix 0.5 mole of each reactant. allow the system to come to equilibrium, and then add another mole of H2 and allow the system to reach equilibrium again. or (b) you mix 1.5 moles of H2 and 0.5 mole of I2 and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.
A sample of gaseous nitrosyl bromide (NOBr) was placed in a container tiued with a frictionless, massless piston, where it decomposed at 25C according to the following equation: 2NOBr(g)2NO(g)+Br2(g) The initial density of the system was recorded as 4.495 g/L. After equilibrium was reached, the density was noted to be 4.086 g/L. a. Determine the value of the equilibrium constant K for the reaction. b. If Ar(g) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of K? Explain each answer
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.

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Write the equilibrium constant expression (Ke) for each of the following reactions. a. 302(9) = 203(9) b. 02(9) =0 2-03(9)