B)
The given reaction is:
2 NO(g) + Br2(g) <-> 2 NOBr(g)
Kc = [NOBr]^2 / [NO]^2[Br2]
= 2.4*10^-2
Now we have:
2 NOBr(g) <—> 2 NO(g) + Br2(g)
Kc’ = [NO]^2[Br2]/[NOBr]^2
= 1/Kc
= 1/(2.4*10^-2)
= 42
Answer: 42
The equilibrium constant for the reaction 2 NO(g) + Br2 = 2 NOBr(9) is Ke = 2.4 x 10-2 at certain temperature. Calc...
The equilibrium constant for the reaction 2 NO(g) + Br2 = 2 NOBr(9) is Ke = 2.2 x 10" at certain temperature. K = 45 Previous Answers Correct Correct answer is shown. Your answer 45.45 was either rounded differes significant figures than required for this part. Part 6 Calculate K. for NOBr(g) - NO(g) + Bra(g). Express your answer using two significant figures. A = 0 ?
Part C NO(g) Br2 (g) Calculate Ke for NOBR(g) Express your answer using two significant figures. ΑΣΦ ? К.- Submit Request Answer
The equilibrium constant, Kc, for the following reaction is 2.0. If the equilibrium mixture contains 2.3 M NO and 0.80 M Br2 , what is the molar concentration of NOBr? 2NOBr(g) = 2NO(g) + Br2 (g) Express your answer to two significant figures and include the appropriate units. Å R 0 2 ? [NOBr] = Value Units Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining
Constants I Periodic Table The following equilibrium exists at 178°C. 2 NO(g) + Bra(g) 2 NOBrg) Calculate the value of K for this reaction if an equilibrium mixture consisting of 4.51 g of NO, 3.68 g of Brą, and 7.4 g of NOBr is present in a 10.0L vessel Express your answer to throw significant figures. | ΑΣφ και K- Submit Request Answer Part B > Calculate the value of K, for this reaction it an equilibrium mixture consisting of...
2. Determine Ke for the reaction (Answer: 9.7x10-16): 2 N2(g) + 1/2O2(g) + Brz(g) $ NOBr (g) From the following information (Hint: Manipulate the equations first, then manipulate K values): 2 NO(g) 5 N2(g) + O2(g) NO (g) + Br2 (g) $ NOBr (g) Ke=2.1x1030 Ko=1.4
For the reaction N2(g) + 3H2(g) = 2NH3(g) what is the value of Ke at 500°C if the equilibrium concentrations are as follows: [H2] = 0.40 M, (N2) = 0.40 M, and (NH3) = 1.9 M Express the equilibrium constant to two significant figures. V AE OE ? Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Set up the equilibrium-constant expression for this reaction using the equilibrium concentrations. The product concentrations a the reactant concentrations are...
At a certain temperature, the equilibrium constant K for the following reaction is 4.3 x 10': NO2(g) + NO(g) = 2 NO2(g) Use this information to complete the following table. There will be very little NO3 and NO. Suppose a 41. L reaction vessel is filled with 0.79 mol of NO2. What can you say about the composition of the mixture in the vessel at equilibrium? There will be very little NO2. x 6 ? O Neither of the above...
Consider the following reaction: 2NO(g)+Br2(g)⇌2NOBr(g) ,Kp=28.4, at 298 K In a reaction mixture at equilibrium, the partial pressure of NO is 102 torr and that of Br2 is 134 torr . Part A What is the partial pressure of NOBr in this mixture? Express your answer in torrs to three significant figures.
The reaction below has an equilibrium constant of Kp = 2.26 x 104 at 298 K. CO(g) + 2H2(g) = CH3OH(9) Part A Calculate Kp for the reaction below. CH3OH(g) = CO(g) + 2H2(g) Express your answer to three significant figures. Ky = 4.42-10-5 Submit Previous Answers Completed Part B Calculate K, for the reaction below. CO(g) + H2(g) = = CH3OH(g) Express your answer to three significant figures. ΟΙ ΑΣΦ ? K = Part C Calculate K, for the...
Consider the following elementary reaction: NO(9) +Br2(9) ► NOBr (9) Suppose we let ky stand for the rate constant of this reaction, and k _ 1 stand for the rate constant of the reverse reaction. Write an expression that gives the equilibrium concentration of Brz in terms of kj, k-], and the equilibrium concentrations of NO and NOBr2. Br2] = 0 0/0 . O. Х 5 ?