Question

For the first order reaction, plot of log₁₀ [A]_t against time 't' is a straight line with a negative slope equal to ____________.

Options

• −2.303 k

• `(-2.303)/"k"`

• `"k"/2.303`

• `2.303/"k"`

Answer 1

For the first order reaction, plot of log₁₀ [A]_t against time 't' is a straight line with a negative slope equal to `underline("k"/2.303)`.

Explanation:

k = `2.303/"t" xx log_10  (["A"]_0)/(["A"]_"t")`

`"kt"/2.303` = log₁₀ [A]₀ − log₁₀ [A]_t

∴ log₁₀ [A]_t = `-"kt"/2.303 + log_10 ["A"]_0` .......[comparing with y = 1mx + c where m = slope]

∴ Slope = `-"k"/2.303`

∴ Negative slope = `"k"/2.303`