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STD 12 - 3. Chemical kinetics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3. Chemical kinetics4 Marks
MCQ
A substance $A$' decomposes by a first order reaction starting initially with $ [A] =2.00\,m $ and after $200\,\min$ $ [A] = 0.15\,m$. For this reaction what is the value of $k$
  • ✓
    $1.29 \times {10^{ - 2}}\,{\min ^{ - 1}}$
  • B
    $2.29 \times {10^{ - 2}}\,{\min ^{ - 1}}$
  • C
    $3.29 \times {10^{ - 2}}\,{\min ^{ - 1}}$
  • D
    $4.40 \times {10^{ - 2}}\,{\min ^{ - 1}}$
✓

Answer

Correct option: A.
$1.29 \times {10^{ - 2}}\,{\min ^{ - 1}}$
a
(a)Given $A(a) = 2.00\,m, $ $ t = 200\, min $ and $a(a-x) = 0.15\,m$

we know

$k = \frac{{2.303}}{t}\log \frac{a}{{a - x}} = \frac{{2.303}}{{200}}\log \frac{{2.00}}{{0.15}}$

$ = \frac{{2.303}}{{200}} \times (0.301 + 0.824) = 1.29 \times {10^{ - 2}}\,{\min ^{ - 1}}$

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