
Calculating the rate of reactionCalculate the rate of reaction, in mol dm-3 s-1, if 0.0440 g of ethyl ethanoate, CH3COOC2H5, (M = 88.0 g mol-1) is formed in 1.00 min from a reaction mixture of total volume 400 cm3
Answer
Step 1: Calculate the amount of ethyl ethanoate in moles:

amount of ethyl ethanoate = 0.0440 g ÷ 88.0 g mol-1
= 0.0005 mol
Step 2: Calculate the concentration of the product:

concentration of ethyl ethanoate = 0.0005 mol ÷ 0.400 dm3
= 0.00125 mol dm-3
Step 3: Calculate the rate:
rate of reaction = 0.00125 mol dm-3÷ 60 s
rate of reaction = 2.08 x 10-5 mol dm-3 s-1

Isomerisation of cyclopropane
Concentrations of Cyclopropane & Propene Table


The graph shows that the concentration of propene increases with time


Line a shows the average rate over the first five minutes whereas line b shows the actual initial rate found by drawing a tangent at the start of the curve. The calculated rates are very similar for both methods

= 0.0009 mol dm-3 s-1


= 0.0005 mol dm-3 s-1

The rate of reaction at three different concentrations of cyclopropane is calculated by drawing tangents at those points in the graph
Change in rate with Increasing Concentration of Cyclopropane


The graph shows a directly proportional correlation between the concentration of cyclopropane and the rate of reaction
To calculate the rate of reaction you can either use the increase in concentration of products (like in the example above) or the decrease in concentration of reactants.
转载自savemyexams
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