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mina [271]
3 years ago
10

Consider the following reaction: NO( g) + SO 3( g) ⇌ NO 2( g) + SO 2( g) A reaction mixture initially contains 0.86 atm NO and 0

.86 atm SO 3. Determine the equilibrium pressure of NO 2 if K p for the reaction at this temperature is 0.0118. 0.048 atm 0.012 atm 0.85 atm 0.084 atm 0.78 atm
Chemistry
1 answer:
DIA [1.3K]3 years ago
5 0

Answer:

The equilibrium pressure of NO2 is 0.084 atm

Explanation:

Step 1: Data given

A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3.

Kp = 0.0118

Step 2: The balanced equation

NO( g) + SO3( g) ⇌ NO2( g) + SO2( g)

Step 3: The initial pressures

p(NO) = 0.86 atm

p(SO3) = 0.86 atm

p(NO2) = 0 atm

p(SO2) = 0 atm

Step 4: The pressure at the equilibrium

For 1 mol NO we need 1 mol SO3 to produce 1 mol NO2 and 1 mol SO2

p(NO) = 0.86 -x atm

p(SO3) = 0.86 -xatm

p(NO2) = x atm

p(SO2) = x atm

Step 5: Define Kp

Kp = ((pNO2)*(pSO2)) / ((pNO)*(pSO3))

Kp = 0.0118 = x²/(0.86 - x)²

X = 0.08427

p(NO) = 0.86 -0.08427 = 0.77573 atm

p(SO3) = 0.86 -0.08427 = 0.77573 atm

p(NO2) = 0.08427 atm

p(SO2) = 0.08427 atm

The equilibrium pressure of NO2 is 0.08427 atm ≈ 0.084 atm

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A 2.5 L flask is filled with 0.25 atm SO3, 0.20 atm SO2, and 0.40 atm O2, and allowed to reach equilibrium. Assume at the temper
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Explanation:

Reaction equation for the given chemical reaction is as follows.

      2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}

Equation for reaction quotient is as follows.

         Q = \frac{P^{2}_{SO_{2}} \times P_{O_{2}}}{P^{2}_{SO_{3}}}

             = \frac{(0.20)^{2} \times 0.40}{(0.25)^{2}}

             = 0.256

As, Q > K (= 0.12)

The effect on the partial pressure of SO_{3} as equilibrium is achieved by using Q, is as follows.

  • This means that there are too much products.
  • Equilibrium will shift to the left towards reactants.
  • More SO_{3} is formed.
  • Partial pressure of SO_{3} increases.
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