Question 35 The initial concentrations of NO, Br2 and NOBr are 2.50 M, 5.00 M and 1.00 M respectively. At equilibrium, the concentration of NO is 2.00 M. Calculate the Kc for the equilibrium system: 2NO(g) + Br₂(g) →→→→→→ 2NOBr(g) A 0.118 B 0.0139 0.0500 0.222 C O D A B OC D

Question 35 The initial concentrations of NO, Br2 and NOBr are 2.50 M, 5.00 M and 1.00 M respectively. At equilibrium, the concentration of NO is 2.00 M. Calculate the Kc for the equilibrium system: 2NO(g) + Br₂(g) →→→→→→ 2NOBr(g) A 0.118 B 0.0139 0.0500 0.222 C O D A B OC D

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 14.95QE: Nitrogen, hydrogen, and ammonia are in equilibrium in a 1000-L reactor at 550 K. The concentration...

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QUESTION#1: TOPIC: Calculating an equilibrium constant from a heterogeneous equilibrium composition Nickel carbonyl decomposes to form nickel and carbon monoxide, like this: NiCO4(g)→Ni(s)+4CO(g) At a certain temperature, a chemist finds that a 2.9 L reaction vessel containing a mixture of nickel carbonyl, nickel, and carbon monoxide at equilibrium has the following composition: compound amount NiCO4 0.792 g Ni 71.9 g CO 1.17 g Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits. Kc= ? QUESTION#2: TOPIC: Calculating the pH of a strong acid solution A chemist dissolves 185. mg of pure hydroiodic acid in enough water to make up 300. mL of solution. Calculate the pH of the solution. Round your answer to 3 significant decimal places. ?
I solved part I but I'm having a hard time with part 2: Question 1. A student mixes 5.00 mL of 2.00 x 10‐3 M Fe(NO3)3 with 5.00 mL 2.00 x 10‐3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN+2 is 1.40 x 10‐4 M. What is the initial concentration in solution of the Fe+3 and SCN‐? What is the equilibrium constant for the reaction? My solution: number of Fe +3 = M*V = 2*10^-3 M * 0.005 L = 1*10^-5 Mnumber of SCN-= M*V = 2*10^-3 M * 0.005 L = 1*10^-5 MTotal volume = 5 +5 = 10 mL = 0.01 LInitial concentration of Fe+3,[Fe+3] = number of moles / total volume =1*10^-5 / 0.01 =1*10^-3 MInitial concentration of SCN-,[SCN-] = number of moles / total volume =1*10^-5 / 0.01 =1*10^-3 M-------------------------------------------------------------------------------Fe+3 + SCN - --->FeSCN+210^-3 10^-3 0 (initial)10^-3-x 10^-3-x x…
QUESTION 12 Consider the following reaction: CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g)A reaction mixture initially contains 0.50 M CH4 and 0.75 M H2S. If the equilibrium concentration of H2 is 0.44 M, find the equilibrium constant (Kc) for the reaction. A 0.34 B 2.9 C 0.23 D 10. E 0.038
< Progress:0/9 groups Due Apr 18 at 11:55 PM Use the References to access important values if needed for this question. In an aqueous chloride solution cobalt(II) exists in equilibrium with the complex ion CoCl42-. Co2+(aq) is pink and CoCl42-(aq) is blue. ... At Low Temperature the pink color predominates. ... ... At High Temperature the blue color is strong. ... If we represent the equilibrium as:...Co2+(aq) + 4Cl-(aq) CoCl42-(aq)We can conclude that: fill in the blank 1 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. fill in the blank 2 2. When the temperature is decreased the equilibrium constant, K: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question. fill in the blank 3 3. When the temperature is decreased the equilibrium concentration of Co2+: A. Increases B. Decreases…
QUESTION 44 In the fictitious reaction below, the equilibrium concentration of A is 0.2024 M, B is 0.7852 M, and A 2B 3 is 0.3058 M. What is the value of the equilbrium constant k c for this reaction? 2 A (aq) + 3 B (aq) <-> A 2B 3 15.42 1.924 2.856 59.82 0.06485
Kc = 0.040 for the system below at 450 oC. If a reaction is initiated with 0.40 mole of Cl2 and 0.40 mole of PCl3 in a 2.0 liter container, what is the equilibrium concentration of Cl2 in the same system? PCl5(g) ⇌ PCl3(g) + Cl2(g) This question was answered but I coudn't read the solution. It was all jumbled up. I've included a picture of the answered question below for review. Can you please make it more coherent?
For the hypothetical reactions 1 and 2, K1 = 102 and K2 = 10–4. 1. A2(g) + B2(g) ↔ 2AB(g) 2. 2A2(g) + C2(g) ↔ 2A2C(g) 3. A2C(g) + B2(g) ↔ 2AB(g) + (1/2)C2(g) What is the value for K for reaction 3? Question 21 options: A) 10–2 B) 104 C) 106 D) 102 E) 10–4
The following system was set up and allowed to come to equilibrium: 2 NO2(g) ⇌ N2O4 At equilibrium, the concentration of nitrogen dioxide was 2.45 mol/L, and the value of the equilibrium constant for the reaction at that temperature is 0.835. What was the concentration of the dinitrogen tetraoxide? Answer in mol/L to 3 decimal places.
Question 52 The reaction 2HCl(g) ⇌ H2(g) + Cl2(g) has Kc = 3.2 × 10−34 at 25 °C. If the reaction vessel initially contains 0.0750 mol L−1 of HCl, what will be the concentrations of H2 and Cl2 at equilibrium?
question 14 The equilibrium constant, Kc, for the following system is 4.62. 2SO2(g) + O2(g) ⇌ 2SO3(g) A system was prepared to give these initial concentrations, [SO3] = 0.015 M [O2] = 0.00855 M, [SO2] = 0.500 M. In which direction will the reaction go and why? Question 14 options: The reaction goes to the right, Qc > Kc The reaction goes to the left, Qc < Kc The reaction goes to the right, Qc < Kc The reaction goes to the left, Qc > Kc It depends on the pressure of the system. The equilibrium position will remain the same.
At 2257 K and 1 atm total pressure, water is 1.77 % dissociated at equilibrium by the reaction 2 H2O = 2H2(g) +O2 (g) Calculate (a) K; (b) △Go at this temperature. Answer: [2.85 x 10-6 ; + 239.6 kJ mol-1 ]
K=4.9x10-3 for the reaction CO2(g)+H2(g)⇄ CO(g)+H2O(g). If [CO2]=0.00025 M, [H2]=0.0010 M, & [CO]=0.000015 M, what is the concentration at equilibrium of H2O? Answer 8.2x10-5M I got 1.84x10^-14? Can someone explain this to me please.

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