1: A 100 L reaction container is charged with 0.730 mol of NOBr, which decompose

Question

1: A 100 L reaction container is charged with 0.730 mol of NOBr, which decomposes at a certain temperature** (say between 100 and 150 oC) according to the following reaction: NOBr(g) NO(g) + 0.5Br2(g) At equilibrium the bromine concentration is 2.27x10-3 M. Calculate Kc (in M0.5) **Not specifying the temperature allows for a more liberal use of random numbers.

2: Consider the formation of hydrogen fluoride:

H2(g) + F2(g) 2HF(g)

If a 1.3 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0097 M H2 is connected to a 4.7 L container filled with 0.033 M F2. The equilibrium constant, Kp, is 7.8 x 1014 (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium.

3: Suppose a 4.00 L nickel reaction container filled with 0.0071 M H2 is connected to a 4.00 L container filled with 0.155 M F2. Calculate the molar concentration of H2 at equilibrium.

In depth explanations would be VERY appreciated, I feel like iI'm lost with this topic, will rate highly!

Explanation / Answer

1. NOBr (g) NO(g) + 0.5 Br2(g)

initial concetration 0.730 mol 0 0

equilibrium conc 0.73 - x x 0.5 x

since, 0.5x = 2.27 * 10-3 M = [Br2]

x = 4.54 * 10-3 M = [NO]

[NOBr] =  0.73 - x = 0.73 - 4.54 * 10-3 = 0.72 M

now, Kc = [NO]  [Br2 ] [NOBr] = 4.54 * 10-3 M * 2.27 * 10-3 M   0.72 M

Kc = 14.3* 10-6

2. H2(g) + F2(g) 2 HF(g)

calculate no .of moles of H2 & F2 first

[H2] = no. of moles = Molarity * volume of solution in L = 0.0097 M * 1.3 L = 0.01 moles

[F2 ] = no. of moles = Molarity * volume of solution in L = 0.033 M * 4.7 L = 0.15 moles

H2(g) + F2(g) 2 HF(g)

initial conc 0.01 mole 0.15 mole 0

equlb conc 0.01-x 0.15 -x 2x

Kp = [HF]2  [H2] [F2]

Kp = 7.8 x 1014

[HF] >[H2] [F2]

thus, Kp = [HF]2

7.8 x 1014 = [2x]2

4x2 = 7.8 x 1014  

x= 1.39 * 107 mole =  [HF]

3. H2(g) + F2(g) 2 HF(g)

calculate no .of moles of H2 & F2 first

[H2] = no. of moles = Molarity * volume of solution in L = 0.0071M * 4 L = 0.028 moles

[F2 ] = no. of moles = Molarity * volume of solution in L = 0.155 M * 4 L = 0.038 moles

H2(g) + F2(g) 2 HF(g)

initial conc 0.028 moles 0.038 moles 0

eqlb conc 0.028-x 0.038-x 2x

[ H2 ] = 0.028-x

since x calculated in answer 2 is loo large than 0.028 ; we can say at equilibrium conc. of H2 is negligible

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