The first order rate constant for the composition of CaCO3 at 700 K is 6.36 × 10–3 s–1 and activation energy is 209 kJ mol–1. Its rate constant (in s–1) at 600 K is x × 10–6. The value of x is _____.

Question

The first order rate constant for the composition of CaCO3 at 700 K is 6.36 × 10–3 s–1 and activation energy is 209 kJ mol–1. Its rate constant (in s–1) at 600 K is x × 10–6. The value of x is _____. (Nearest integer) [Given R = 8.31 J K–1 mol–1 ; log 6.36 × 10–3 = –2.19, 10–4.79 = 1.62 ×10–5] 700 K ij CaCO3 ds izFke dksfV ds vi?kVu ds fy, osx fLFkjkad 6.36 × 10–3 s–1 gS] rFkk lfØ;.k ÅtkZ 209 kJ mol–1 gSA 600 K ij bldk osx fLFkjkad (s–1 esa) x × 10–6 gSA x dk eku gS _____ A (fudVre iw.kk±d esa) [fn;k gS % R = 8.31 J K–1 mol–1 ; log 6.36 × 10–3 = –2.19, 10–4.79 = 1.62 ×10–5]

Prashant Kotian · Accepted Answer

Correct answer: .. K700 = 6.36 × 10–3 s–1 ; K600 = x × 10–6 s–1 Ea = 209 kJ/mol Applying ; log         2 T T K K =         2 a T T R . E log       700 K K =         1 1 R . Ea log           3 K 36 . =           700 31 . 303 . 1000 log(6.36 × 10–3) – logK600 = 2.6  logK600 = –2.19 – 2.6 = –4.79  K600 = 10–4.79 = 1.62 × 10–5 = 16.2 × 10–6 = x × 10–6  x = 16 | JEE MAIN-2021 | DATE : 27-08-2021 (SHIFT-2) | PAPER-1 | OFFICAL PAPER | CHEMISTRY PAGE # 10

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