HCO2H(g) �¨ CO2(g) + H2(g) The half-life for the reaction at 550 �‹C is 24 seconds. How many seconds does it take for the formic acid concentration to decrease by 75%?

If a movie has 0.16 hour remaining, how many seconds of movie remain?

To determine the time it takes for the formic acid concentration to decrease by 75%, we need to use the concept of half-life.

First, let's understand what half-life means in this context. The half-life of a reaction is the time it takes for the concentration of a reactant to decrease by half. In this reaction, the half-life is given as 24 seconds.

Since we want to find the time it takes for the formic acid concentration to decrease by 75%, we first need to find the number of half-lives it would take for the concentration to decrease by 75%.

To calculate this, we can use the formula:

Number of half-lives = (log(final concentration / initial concentration)) / (log(1/2))

In this case, the final concentration is 25% (0.25) of the initial concentration because we want to determine the time it takes for the concentration to decrease by 75%.

Let's calculate the number of half-lives:

Number of half-lives = (log(0.25) / log(1/2))

Number of half-lives = (log(0.25) / log(0.5))

Using a calculator, we find:

Number of half-lives ≈ 2.3219

Now we have the number of half-lives it takes for the formic acid concentration to decrease by 75%. To find the time it takes, we need to multiply this number by the half-life value:

Time = Number of half-lives * Half-life

Time = 2.3219 * 24 seconds

Time ≈ 55.73 seconds

Therefore, it takes approximately 55.73 seconds for the formic acid concentration to decrease by 75% at 550 �‹C.

.25=e^(-.692t/24)

take the ln of each side, solve for t. You need to do that calc for practice.

Now, the easy way with the numbers given.
In the first half life, you have .50 left
in the second half life, you have .25 left....which means .75 has been consumed.
two half lives...