Starting with 0.280 mol SbCl3 and 0.160 mol Cl2, how many moles of SbCl5, SbCl3, and Cl2 are present when equilibrium is established at 248°C in a 2.50 L flask? Kc = 2.5 x 10^-2 at... Starting with 0.280 mol SbCl3 and 0.160 mol Cl2, how many moles of SbCl5, SbCl3, and Cl2 are present when equilibrium is established at 248°C in a 2.50 L flask? Kc = 2.5 x 10^-2 at 248°C. Read more

Steps to Solve

1. Write the equilibrium expression

For the reaction ( \text{SbCl}_5(g) \rightleftharpoons \text{SbCl}_3(g) + \text{Cl}_2(g) ), the equilibrium constant expression ( K_c ) is given by:

$$ K_c = \frac{[\text{SbCl}_3][\text{Cl}_2]}{[\text{SbCl}_5]} $$

2. Convert initial amounts to concentrations

Using the initial moles and the volume (2.50 L), calculate the initial concentrations:

• Initial concentration of ( \text{SbCl}_3 ):

$$ [\text{SbCl}_3] = \frac{0.280 \text{ mol}}{2.50 \text{ L}} = 0.112 , \text{M} $$

• Initial concentration of ( \text{Cl}_2 ):

$$ [\text{Cl}_2] = \frac{0.160 \text{ mol}}{2.50 \text{ L}} = 0.064 , \text{M} $$

• Initial concentration of ( \text{SbCl}_5 ):

$$ [\text{SbCl}_5] = 0 , \text{M} $$ (since no ( \text{SbCl}_5 ) is initially present)

3. Set up the change in concentrations

Let ( x ) be the change in concentration of ( \text{SbCl}_5 ) at equilibrium. The equilibrium concentrations will be:

• ( [\text{SbCl}_3] = 0.112 - x )
• ( [\text{Cl}_2] = 0.064 - x )
• ( [\text{SbCl}_5] = x )

4. Substitute into the equilibrium expression

Substituting the equilibrium concentrations into the equilibrium expression:

$$ 2.5 \times 10^{-2} = \frac{(0.112 - x)(0.064 - x)}{x} $$

5. Solve the equation

To solve for ( x ), rearrange and simplify:

$$ 2.5 \times 10^{-2} x = (0.112 - x)(0.064 - x) $$

Expand the right side:

$$ 2.5 \times 10^{-2} x = 0.007168 - 0.112x - 0.064x + x^2 $$

Combine like terms and rearrange into a standard quadratic form:

$$ x^2 - (0.176 + 2.5 \times 10^{-2})x + 0.007168 = 0 $$

6. Use the quadratic formula

Using the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), where ( a = 1, b = -(0.176 + 2.5 \times 10^{-2}), c = 0.007168 ):

• Calculate ( b^2 - 4ac ) and solve for ( x ).

7. Determine equilibrium moles

Once ( x ) is calculated, find the equilibrium concentrations:

• Calculate moles of ( \text{SbCl}_5, \text{SbCl}_3, \text{Cl}_2 ) using:

$$ \text{Moles} = [\text{Concentration}] \times \text{Volume (L)} $$