The decomposition of formic acid follows first-order kinetics.
HCO₂H(g) $\to$ CO₂(g) + H₂(g)
The half-life for the reaction at 550 ^$\circ$C is 24 seconds.How many seconds does it take for the formic acid concentration to decrease by 87.5%?

Question

The decomposition of formic acid follows first-order kinetics.
HCO₂H(g)

\to

CO₂(g) + H₂(g)
The half-life for the reaction at 550 ^$\circ$C is 24 seconds.How many seconds does it take for the formic acid concentration to decrease by 87.5%?

Quizplus · Accepted Answer

The rate constant (k) can be calculated using the half-life equation:

t1/2 = ln(2)/k
24 = ln(2)/k

k = ln(2)/24

To calculate the time it takes for the concentration to decrease by 87.5%, we can use the first-order integrated rate law equation:

ln([A]/[A]0) = -kt

where [A] is the concentration at a given time, [A]0 is the initial concentration, k is the rate constant, and t is time.

We can rearrange the equation to solve for t:

t = -ln([A]/[A]0)/k

When the concentration is decreased by 87.5%, it is at 12.5% of the original concentration ([A]/[A]0 = 0.125).

t = -ln(0.125)/k
t = 72 seconds

Therefore, the answer is D) 72 s.

The Decomposition of Formic Acid Follows First-Order Kinetics →\to→ CO2(g)+ H2(g) the Half-Life for the Reaction at 550

The Decomposition of Formic Acid Follows First-Order Kinetics $\to$ CO₂(g)+ H₂(g)
the Half-Life for the Reaction at 550