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2 years ago

Balance each reaction and write its reaction quotient, Qc:(b) POCl₃(g) ⇄ PCl₃(g) + O₂(g)

Chemistry

1 answer:

Komok [63]2 years ago

4 0

When we balance the given equation

POCl₃(g) ⇄ PCl₃(g) + O₂(g)

We will get

2POCl₃(g) ⇄ 2PCl₃(g) + O₂(g)

Solution:

Balancing the given equation

POCl₃(g) ⇄ PCl₃(g) + O₂(g)

Balancing the number of O

2POCl₃(g) ⇄ PCl₃(g) + O₂(g)

Balancing the number of P and Cl

2POCl₃(g) ⇄ 2PCl₃(g) + O₂(g)

We get the balanced equation

2POCl₃(g) ⇄ 2PCl₃(g) + O₂(g)

The reaction quotient will be

Qc = [product] / [reactant]

Qc = [PCl₃(g) + O₂(g)] / [POCl₃(g) ]

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A mixture initially contains AA, BB, and CC in the following concentrations: [A][A]A_1 = 0.550 MM , [B][B]B_1 = 1.40 MM , and [C

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Answer:

The value of the equilibrium constant KC is 1.244

Explanation:

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C

At equilibrium, [A] = 0.430 M and [C] = 0.720 M. Calculate the value of the equilibrium constant, Kc

Step 1: The balanced equation

A+2B<->C

Step 2: The initial concentrations

[A] = 0.550 M

[B]= 1.40 M

[C] = 0.600 M

Step 3: The concentraions at equilibrium

[A] = 0.550 -X = 0.430 M

[B]= 1.40 -2X M

[C] = 0.600 + X = 0.720 M

X = 0.120 M

[A] = 0.550 - 0.120 = 0.430 M

[B]= 1.40 -2*0.120 = 1.16 M

[C] = 0.600 + 0.120 = 0.720 M

Step 4: Calculate Kc

Kc = [C] / [A][B]²

Kc = 0.720 / (0.430*1.16²)

Kc = 1.244

The value of the equilibrium constant KC is 1.244

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