A student ran the following reaction in the laboratory at 431 K: PCI 5(9) PC13(g) + Cl₂(9) When she introduced 2.02 moles of PCI5(g) into a 1.00 liter container, she found the equilibrium concentration of Cl₂(g) to be 3.08x10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.

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A student ran the following reaction in the laboratory at 431 K:

\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

When she introduced 2.02 moles of PCl₅(g) into a 1.00 liter container, she found the equilibrium concentration of Cl₂(g) to be \( 3.08 \times 10^{-2} \, M \).

Calculate the equilibrium constant, \( K_c \), she obtained for this reaction.

\[ K_c = \] 

(Note: The problem ends here, asking for the calculation of the equilibrium constant, \( K_c \), without providing the final calculated value.)
Transcribed Image Text:A student ran the following reaction in the laboratory at 431 K: \[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \] When she introduced 2.02 moles of PCl₅(g) into a 1.00 liter container, she found the equilibrium concentration of Cl₂(g) to be \( 3.08 \times 10^{-2} \, M \). Calculate the equilibrium constant, \( K_c \), she obtained for this reaction. \[ K_c = \] (Note: The problem ends here, asking for the calculation of the equilibrium constant, \( K_c \), without providing the final calculated value.)
Expert Solution
Step 1 Objective

Given, the equilibrium reaction as follows,

PCl5(g) PCl5(g)+Cl2 (g)

with initial amount of PCl5 and equilibrium concentration of chlorine in one liters container.

We are asked to calculate the 

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