CBSE Class 12 · 2024 · Chemical KineticsShort Answer

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Show that in the case of a first order reaction, the time taken for completion of 99% of the reaction is twice the time required for 90% completion of the reaction. (log 10 = 1)

Answer: t₉₉% = (2.303/k)·2 and t₉₀% = (2.303/k)·1, so t₉₉% = 2 t₉₀%.

Solution

For a first order reaction:

t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ]}

For 99% completion,

[R] = [R]_{0} /100

t_{99%} = \frac{2.303}{k} lo g 100 = \frac{2.303}{k} \times 2

For 90% completion,

[R] = [R]_{0} /10

t_{90%} = \frac{2.303}{k} lo g 10 = \frac{2.303}{k} \times 1

Dividing:

t_{99%} = 2 t_{90%}

CBSE Class 12 Board · 2024 · Subjective

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