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Formic acid decomposes at 550°C according to the following reaction: HCO2H (g) → CO2...
Formic acid decomposes at 550°C according to the following reaction:
HCO2H (g) → CO2 (g) + H2 (g)
The reaction follows first-order kinetics. In an experiment, it is determined that 75% of a sample of HCO2H has decomposed in 72 seconds. Determine t1/2 for this reaction.
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For first order processes, we have two equations that we can use. The first is the integrated rate law
ln([At]/[Ao] = - kt
Where At is the concentration at time t, Ao is the initial concentration, k is the rate constant and t is the time.
Since we are given percentages, we can use that value as the concentrations in the formula because what we are really worried about is the ratio. The initial concentration will be 100%. If we decompose 75% of the sample over some time period, then we are left with just 25% of the sample so that will be [At]. Now, we can plug in what we know and solve for k
ln (25/100) = - k (72 sec)
k = 0.0193 sec^-1
Now, we can plug that into our half life formula
t1/2 = 0.693/k
t1/2 = 0.693/0.0193
t1/2 = 36 seconds
Since the time in the integrated rate law was seconds, k has units of inverse seconds and the half-life will be in units of seconds.
Posted by mlsiasebs on April 26, 2012 at 4:14 PM (Answer #1)
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