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The equilibrium constant Kc for the reaction 2NO(g)+O2(g)⇌2NO2(g) is 6.9×105 at 500 K. A 5.0 L reaction vessel at 500 K was filled with 0.056 mol of NO, 1.0 mol of O2, and 0.90 mol of NO2. Part A What is the value of Qc?

Respuesta :

Answer:

Q = 1291.5

Explanation:

Step 1: Data given

Kc = 6.9 * 10^5 at 500 K

Volume = 5.0 L

Number of moles NO = 0.056 moles

Number of moles O2 = 1.0 moles

Number of moles NO2 = 0.90 moles

Step 2: The balanced equation

2NO(g) + O2(g) ⇌ 2NO2(g)

Step 3: Calculate the concentration

Concentration = moles / volume

[NO] = 0.056 moles / 5.0 L

[NO]= 0.0112 M

[O2] = 1.0 mol / 5.0L

[O2] = 0.2 M

[NO2] = 0.90 moles / 5.0L

[NO2] = 0.18 M

Step 4: Calculate Q

Q = [NO2]²/[NO]²[O2]

Q = [0.18²]/[(0.0112²)(0.2)]

Q = 1291.5

jayilych4real

The value of Qc is 1291.5

The Calculation and Parameters:

Based on the available data, we can get

Kc = 6.9 * 10^5 at 500 K

  • Volume = 5.0 L
  • Number of moles NO = 0.056 moles
  • Number of moles O2 = 1.0 moles
  • Number of moles NO2 = 0.90 moles

Next, we have to balance the equation:

  • 2NO(g) + O2(g) ⇌ 2NO2(g)

Next, we calculate the concentration

Concentration = moles / volume

  • [NO] = 0.056 moles / 5.0 L
  • [NO]= 0.0112 M
  • [O2] = 1.0 mol / 5.0L
  • [O2] = 0.2 M
  • [NO2] = 0.90 moles / 5.0L
  • [NO2] = 0.18 M

Finally, we calculate for Q

  • Q = [NO2]²/[NO]²[O2]
  • Q = [0.18²]/[(0.0112²)(0.2)]
  • Q = 1291.5

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