Question

The thermal decomposition of HCO2H is a first order reaction with a rate constant of 2.4 × 10-3 s-1 at a certain temperature. Calculate how long will it take for three-fourths of initial quantity of HCO2 H to decompose. (log 0.25 = -0.6021)

Answer 1

Explanation:

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Answer 2

Given:

The thermal decomposition of HCO2H is a first order reaction with a rate constant of 2.4 × 10-3 s-1 at a certain temperature.

To Find :

How long will it take for three-fourths of initial quantity of HCO2 H to decompose.

Solution:

For a first order reaction

t = 2.303/k [ log ( R./R ) ]

Where ,

t = time

k =  Rate constant

R = instantaneous concentration of reactant

R. = Initial concentration

For,  k = 2.4× $10^{-3}$ $s^{-1}$

R = R./4

t = $\frac{2.303}{ 2.4 * 10^{-3} }$ [ log ( R./{R./4}) ]

t = $\frac{2.303}{ 2.4 * 10^{-3} }$ [ log ( 4 ) ] ___________(1)

Also, log(1/4) = - log(4)

log(4) = - log(1/4)

log(4) = 0.6021 _______(2)

Putting (2) in (1)

t =  $\frac{2.303}{ 2.4 * 10^{-3} }$ [0.6021]

t = 595. 03 s

Hence , Time required will be t = 595. 03 s