The thermal decomposition of HCO$_2$H is a first order reaction with a rate constant of $2.4\times10^{-3}$ s$^{-1}$ at a certain temperature. Calculate how long it will take for three-fourths of the initial quantity of HCO$_2$H to decompose. (log 0.25 = −0.6021)

CBSE Class 12 · 2011 · All India · Chemical KineticsShort Answer

The thermal decomposition of HCO

_{2}

H is a first order reaction with a rate constant of

2.4 \times 1 0^{- 3}

^{- 1}

at a certain temperature. Calculate how long it will take for three-fourths of the initial quantity of HCO

_{2}

H to decompose. (log 0.25 = −0.6021)

Answer: t ≈ 578 s

Three-fourths decomposed means

[R] = [R]_{0} /4

t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ] _{0} /4} = \frac{2.303}{2.4 \times 1 0 ^{- 3}} lo g 4

= \frac{2.303 \times 0.6021}{2.4 \times 1 0 ^{- 3}}

t = 578 s (\approx 9.6 min)

CBSE Class 12 Board · 2011 · All India · Subjective