Numbers 6 and 7 please When the following reaction reaches equilibrium, A(g) + 2

Question

Numbers 6 and 7 please

When the following reaction reaches equilibrium, A(g) + 2 B(g) Reversible C(g) the following concentrations are measured: [A] = 0.60; [B] = 0.20; [C] = 0.55. What is the value of K_c for this reaction An equilibrium mixture of PCl_5, PCl_3,and Cl_2, at a certain temperature in a 5.0 L container consists of 0.80 mole PCl_5, 0.55 mole PCl_3, and 1.2 mole Cl_2. Calculate K_c for the reaction: PCl_3(g) + Cl_2(g) Reversible PCl_5(g) Calculate K_c for the reaction: 3H_2(g) + N_2(g) Reversible 2NH_3(g) given that the equilibrium concentrations are: [H_2] = 1.5; [NH_3] = 0.24; [N_2] = 2.5.

Explanation / Answer

6.

A + 2B => C

equilibrium constant Kc= [C]/([A][B]2)

[A]= 0.6

[B]=0.2

[C]= 0.55

plugging in the values;

Kc= 22.9167

7.

PCl3 + Cl2 => Pl5

calculating molarity from moles

[PCl3]= 0.55/5= 0.11 M

[PCl5]= 0.8/5 = 0.16 M

[Cl2]= 1.2/5= 0.24

now, Kc=[PCl5]/[PCl3][Cl2]

pluggging in the values,

we get Kc= 6.06

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