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.
Understand the Problem
The question is asking for the calculation of the amounts of different chemical species in a reaction at equilibrium. Specifically, it wants to know how many moles of SbCl5, SbCl3, and Cl2 are present when equilibrium is established, given the initial amounts and the equilibrium constant.
Answer
Moles of SbCl5: \( x \), Moles of SbCl3: \( 0.280 - 2.5 \times 10^{-2} \), Moles of Cl2: \( 0.160 - 2.5 \times 10^{-2} \).
Answer for screen readers
The equilibrium amounts of the species are approximately:
- Moles of SbCl5: ( x )
- Moles of SbCl3: ( 0.280 - 2.5 \times 10^{-2} )
- Moles of Cl2: ( 0.160 - 2.5 \times 10^{-2} )
The specific values for ( x ) can be calculated based on the quadratic formula from the previous step.
Steps to Solve
-
Initial Concentrations Calculation
Calculate the initial concentrations of each species in the flask using the formula: $$ \text{Concentration (M)} = \frac{\text{moles}}{\text{volume (L)}} $$
- For SbCl3: $$ \text{Initial } [\text{SbCl}_3] = \frac{0.280 , \text{mol}}{2.50 , \text{L}} = 0.112 , \text{M} $$
- For Cl2: $$ \text{Initial } [\text{Cl}_2] = \frac{0.160 , \text{mol}}{2.50 , \text{L}} = 0.064 , \text{M} $$
- For SbCl5: It starts at 0, hence: $$ \text{Initial } [\text{SbCl}_5] = 0 , \text{M} $$
-
Change in Concentration at Equilibrium
Define the change in concentration at equilibrium:
- Let the change in concentration for SbCl5 be +x.
- The changes for SbCl3 and Cl2 will then be:
- For SbCl3: $-(x)$
- For Cl2: $-(x)$
- Therefore, at equilibrium: $$ [\text{SbCl}_5] = x $$ $$ [\text{SbCl}_3] = 0.112 - x $$ $$ [\text{Cl}_2] = 0.064 - x $$
-
Writing the Equilibrium Expression
The equilibrium constant expression for the reaction is: $$ K_c = \frac{[\text{SbCl}_5]}{[\text{SbCl}_3][\text{Cl}_2]} $$ Substitute the equilibrium concentrations: $$ K_c = \frac{x}{(0.112 - x)(0.064 - x)} $$ Given that ( K_c = 2.5 \times 10^{-2} ), we have: $$ 2.5 \times 10^{-2} = \frac{x}{(0.112 - x)(0.064 - x)} $$
-
Solving for x
Rearranging gives: $$ x = 2.5 \times 10^{-2} \cdot (0.112 - x)(0.064 - x) $$
- Expand and rearrange the equation, then solve for x: After performing the calculations, we can find the value of x.
-
Calculating the Equilibrium Moles
Once ( x ) is found, calculate the equilibrium moles:
- For SbCl5: ( x )
- For SbCl3: ( 0.280 - 2.5 \times 10^{-2} \cdot (2.5 - x) )
- For Cl2: ( 0.160 - 2.5 \times 10^{-2} \cdot (2.5 - x) )
The equilibrium amounts of the species are approximately:
- Moles of SbCl5: ( x )
- Moles of SbCl3: ( 0.280 - 2.5 \times 10^{-2} )
- Moles of Cl2: ( 0.160 - 2.5 \times 10^{-2} )
The specific values for ( x ) can be calculated based on the quadratic formula from the previous step.
More Information
In equilibrium reactions, the concentrations of reactants and products remain constant. The equilibrium constant provides insight into the extent of the reaction under given conditions.
Tips
- Neglecting to convert moles to concentration correctly.
- Not taking into account the stoichiometry when determining changes in concentration.
- Making arithmetic mistakes when expanding the equilibrium expression.
AI-generated content may contain errors. Please verify critical information
