Effects of Concentration Table
Equilibrium expression linking the equilibrium concentration of reactants and products at equilibrium
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
Graph showing the effects of adding nitrogen on the concentration of reactants and products in the Haber Process
Graph showing the effects of adding nitrogen on the rate of reaction in the Haber Process
Equation to calculate concentration from number of moles and volume
Calculating Kcof ethanoic acid
Ethanoic acid and ethanol react to form the ester ethyl ethanoate and water as follows:
CH3COOH (I) + C2H5OH (I) ⇌ CH3COOC2H5 (I) + H2O (I)
At equilibrium, 500 cm3 of the reaction mixture contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water.
Use this data to calculate a value of Kc for this reaction.
Answer:
Step 1: Calculate the concentrations of the reactants and products:
Step 2: Write out the balanced symbol equation with the concentrations of each chemical underneath:
Step 3: Write out the equilibrium constant for the reaction:
Step 4: Substitute the equilibrium concentrations into the expression and calculate the answer:
Step 5: Deduce the correct units for Kc:
Calculating Kc of ethyl ethanoate
Ethyl ethanoate is hydrolysed by water:
CH3COOC2H5 (I) + H2O (I) ⇌ CH3COOH (I) + C2H5OH (I)
0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture made up to 1 dm3. At equilibrium 0.0654 mol of water are present. Use this data to calculate a value of Kc for this reaction.
Answer:
Step 1: Write out the balanced chemical equation with the concentrations of beneath each substance using an initial, change and equilibrium table:
Step 2: Calculate the concentrations of the reactants and products:
Step 3: Write the equilibrium constant for this reaction in terms of concentration:
Step 4: Substitute the equilibrium concentrations into the expression:
= 0.28
Step 5: Deduce the correct units for Kc:
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