Question

3. In the presence of excess thiocyanate ion, SCN - , the following reaction is first order in chromium (III) ion, Cr +3, the rate constant is 2 x10 -6 s -1.

Cr +3 + SCN - →Cr(SCN) 2+

a) what is the half-life in hours?

b) How many hours would be required for the initial conc of Cr+3 to decrease to the following values: 25% left, 12.5% left?

Answer 1

Answer on Question #70508 - Chemistry - General Chemistry

Question:

3. In the presence of excess thiocyanate ion, SCN - , the following reaction is first order in chromium (III) ion, Cr +3, the rate constant is 2 x10⁻⁶ s⁻¹.

$$\mathrm{Cr} + 3 + \mathrm{SCN} - \rightarrow \mathrm{Cr}(\mathrm{SCN}) 2 +$$

a) what is the half-life in hours?

b) How many hours would be required for the initial conc of Cr+3 to decrease to the following values: 25% left, 12.5% left?

Solution:

a) The half-life is a time on which the initial population decreased by half of its start value. For the first-order reaction the half-life can be calculated by the following concluding equation:

$$\tau_{1/2} = \frac{\ln 2}{k}$$ $$\tau_{1/2} = \frac{0.693}{2 \cdot 10^{-6} s^{-1}} = 3.465 \cdot 10^{5} s$$

The half-life in hours will be:

$$\tau_{1/2} = \frac{3.5 \cdot 10^{5} s}{360 s} = 96.3 h$$

b) If 25% of Cr+3 left it means that time will be equal to $\tau_{1/4}$:

$$t_{25\%} = 2 \cdot \tau_{1/2} = 2 \cdot 96.3 h = 192.6 h$$

12.5% of initial concentration of the ion left will be equal to $\tau_{1/8}$:

$$t_{12.5\%} = 3 \cdot \tau_{1/2} = 3 \cdot 96.3 h = 288.9 h$$

Answer:

a) $\tau_{1/2} = 96.3 h$

b) $t_{25\%} = 192.6 h, t_{12.5\%} = 288.9 h$

Answer provided by www.AssignmentExpert.com