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Gaseous indium dihydride is formed from the elements at elevated temperature:

ln(g)+H2(g)?lnH2(g)Kp = 1.48 at 973 K

Partial pressures measured in a reaction vessel are:
PIn = 0.0650atm , PH2 = 0.0330atm , PInH2 = 0.0790atm

1. Calculate Qp
2. Determine the direction of reaction to attain equilibrium
3. Determine the equilibrium partial pressure of In
4. Determine the equilibrium partial pressure of H2
5. Determine the equilibrium partial pressure of InH2

Respuesta :

Answer:

For 1: The value of [tex]Q_p[/tex] for above reaction is 36.83

For 2: The value of [tex]Q_p[/tex] for above reaction is 36.83

For 3: The equilibrium partial pressure of Indium is 0.126 atm

For 4: The equilibrium partial pressure of hydrogen gas is 0.094 atm

For 5: The equilibrium partial pressure of Indium dihydrogen is 0.018 atm

Explanation:

We are given:

Partial pressure of Indium gas = 0.0650 atm

Partial pressure of hydrogen gas = 0.0330 atm

Partial pressure of Indium dihydride = 0.0790 atm

The given chemical equation follows:

                      [tex]ln(g)+H_2(g)\rightleftharpoons InH_2(g)[/tex]

Initial:               0.065     0.033           0.079

At eqllm:       0.065-x   0.033-x       0.079+x

  • For 1:

The expression of [tex]Q_p[/tex] for above reaction follows:

[tex]Q_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}[/tex]

Putting values in above equation, we get:

[tex]Q_p=\frac{0.079}{0.065\times 0.033}=36.83[/tex]

Hence, the value of [tex]Q_p[/tex] for above reaction is 36.83

  • For 2:

We are given:

[tex]K_p[/tex] of the reaction = 1.48

There are 3 conditions:

  • When [tex]K_{p}>Q_p[/tex]; the reaction is product favored.
  • When [tex]K_{p}<Q_p[/tex]; the reaction is reactant favored.
  • When [tex]K_{p}=Q_p[/tex]; the reaction is in equilibrium.

As, [tex]Q_{p}>K_p[/tex] for the given reaction, the reaction is reactant favored.

Hence, the reaction proceed in the backward direction to attain equilibrium

  • For 3:

The expression of [tex]K_p[/tex] for above reaction follows:

[tex]K_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}[/tex]

Putting values in above equation, we get:

[tex]1.48=\frac{(0.079+x)}{(0.065-x)\times (0.033-x)}\\\\x=-0.061,0.835[/tex]

Neglecting the value of x = 0.835 because the reaction is going backwards. So, by taking this value, the pressure of the reactants will decrease

So, equilibrium partial pressure of Indium = (0.065 - x) = [0.065 - (-0.061)] = 0.126 atm

  • For 4:

The equilibrium partial pressure of hydrogen gas = (0.033 - x) = [0.033 - (-0.061)] = 0.094 atm

  • For 5:

The equilibrium partial pressure of Indium dihydrogen = (0.079 + x) = [0.079 + (-0.061)] = 0.018 atm

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