Consider the following reaction where K_c = 1.20 times 10^-2 at 500 K: Use the R

Question

Consider the following reaction where K_c = 1.20 times 10^-2 at 500 K: Use the References to access important values if needed for this question. PCI_5(g) rightarrow leftarrow PCI_3(g) + CI_2(g) A reaction mixture was found to contain 0.100 moles of PCI_5 (g), 23.03 times 10^-2 moles of PCI_3 (g), and 3.76 times 10^-2 moles of CI_2 (g), in a 1.00 liter container. In order to reach equilibrium PCI_5(g) must be consumed. In order to reach equilibrium K_c must decrease. In order to reach equilibrium PCI_3 must be consumed. Q_c is less than K_c. The reaction is at equilibrium. No further reaction will occur.

Explanation / Answer

PCl5 (g) <----> PCl3 (g) + Cl2 (g)

Kc = 1.20 * 10^-2

Since volume is 1.00 L, moles is equal to molarity

So [PCl5] = 0.100 M, [PCl3] = 2.30 *10^-2 M; [Cl2] = 3.76 * 10^-2 M

Qc = [PCl3] [Cl2] / [PCl5]

= 2.30 * 10^-2 * 3.76 * 10^-2 / (0.100 M)

= 8.65 * 10^-3

This is lower than the value of Kc. So the reaction has to go to the right to reach equilibrium.

1. PCl5 must be consumed for the reaction to go to the right. So it is true

2. Kc is constant. So it is false

3. PCl3 must be formed. So this is false

4. QC is less than Kc as calculated above. So this is true

5. The reaction is not at equilibrium. So this is false.

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