For first order reaction, t=k2.303loga−xa Given : k=2.303×10−3 s−1,a=40 g,a−x=10 g On substituting the given values in Eq. (i), we get t=2.303×10−32.303log1040=103log22=2×103×log2=2×103×0.3010=602 s Alternative method For first order reaction, t1/2=k0.693 t1/2(t50%)=2.303×10−30.693=301 s Also, t75%=2t50% ∴t75%=2×301=602 s
A first order reaction has a rate constant of 2.303 × 10^-3 ~s^-1. The time required for 40 ~g of this reactant to reduce to 10 ~g will be [Given…
NEET UG 2019 — Chemistry Physical Chemistry
A first order reaction has a rate constant of 2.303×10−3 s−1. The time required for 40 g of this reactant to reduce to 10 g will be [Given that log102=0.3010 ]
Held on 30 Apr 2019 · Verified 9 Jul 2026.
Options
- A
230.3 s
- B
301 s
- C
2000 s
- D
602 s
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