The rate of a first order reaction is $0.04 M . S^{- 1}$ at 10 min and $0.03 M . S^{- 1}$ at 20 min after initiation rate constant value is $y \times 10^{- 4} \min^{- 1} .$ Find value of $(y - 0.875) ? (\log_{10}^{4} = 0.6020, {\log_{10}}^{3} = 0.477)$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 287.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\frac{r_{1}}{r_{2}} = \frac{C_{1}}{C_{2}} (∵ n = 1^{s t} o r d e r r \infty c)$

$\frac{γ_{1}}{γ_{2}} = \frac{C_{1}}{C_{2}} = \frac{0.04}{0.03} \Rightarrow K = \frac{2.303}{t_{20} - t_{10}} \log \frac{C_{1}}{C_{2}}$

$∴ K = \frac{2.303}{10} \log \frac{4 \times 10^{- 2}}{3 \times 10^{- 2}} \Rightarrow K = 287 \times 10^{- 4} \min^{- 1}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE