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The equilibrium constant for the decomposition of PCl5 at 250 celcius is 1.05. PCl5(g)-->PCl3(g)+Cl2(g) If the equilibrium pressures of PCl5 and PCl3 are .875 atm and .463 atm, respectively, then what is the equilibrium partial pressure of Cl2 at 250 celsius.

Respuesta :

Answer : The equilibrium partial pressure of Cl₂ at 25°C is, 1.98 atm

Explanation :

For the given chemical reaction:

[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex]

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

[tex]K_p=\frac{P_{PCl_3}\times P_{Cl_2}}{P_{PCl_5}}[/tex]

We are given:

[tex]P_{PCl_5}=0.875atm[/tex]

[tex]P_{PCl_3}=0.463atm[/tex]

[tex]K_p=1.05[/tex]

Putting values in above equation, we get:

[tex]1.05=\frac{0.463\times P_{Cl_2}}{0.875}\\\\P_{Cl_2}=1.98atm[/tex]

Thus, the equilibrium partial pressure of Cl₂ at 25°C is, 1.98 atm

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