Question
The rate law for the decomposition of phosphine $\left(\mathrm{PH}_{3}\right)$ is $$\text { Rate }=-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right]$$ It takes $120 .$ s for $1.00 M$ PH $_{3}$ to decrease to 0.250 M. How much time is required for $2.00\space \mathrm{M} \mathrm{PH}_{3}$ to decrease to a concentration of $0.350\space \mathrm{M} ?$
Step 1
00 M to 0.250 M. This corresponds to two half-lives, because in each half-life, the concentration of a substance decreases by half. Therefore, one half-life is 120 seconds / 2 = 60 seconds. Show more…
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The rate law for the decomposition of phosphine (PH_ $_{3} )$ is $$ \text {Rate} =-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right] $$ It takes $120 .$ s for 1.00$M \mathrm{PH}_{3}$ to decrease to 0.250 M. How much time is required for 2.00$M \mathrm{PH}_{3}$ to decrease to a concentration of 0.350 $\mathrm{M}$ ?
The rate law for the decomposition of phosphine $\left(\mathrm{PH}_{3}\right)$ is $$ \text { Rate }=-\frac{d\left[\mathrm{PH}_{3}\right]}{d t}=k\left[\mathrm{PH}_{3}\right] $$ It takes $120 .$ s for the concentration of $1.00 \mathrm{M} \mathrm{PH}_{3}$ to decrease to $0.250 \mathrm{M} .$ How much time is required for $2.00 \mathrm{M} \mathrm{PH}_{3}$ to decrease to a concentration of $0.350 \mathrm{M} ?$
The rate law for the decomposition of phosphine (PH3) is Rate = k[PH3]. It takes 125 s for 1.00 M PH3 to decrease to 0.200 M. How much time is required for 2.00 M PH3 to decrease to a concentration of 0.350 M? t =
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