Question

For the zero-order decomposition of HI on a gold surface $$ \mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{I}_{2}(g) $$ it takes \(16.0 \mathrm{~s}\) for the pressure of HI to drop from 1.00 atm to 0.200 atm. (a) What is the rate constant for the reaction? (b) How long will it take for the pressure to drop from 0.150 atm to 0.0432 atm? (c) What is the half-life of HI at a pressure of 0.500 atm?

Short answer

Question: Calculate the rate constant for a zero-order decomposition reaction. Then, determine how long it takes for the pressure to drop from 0.150 atm to 0.0432 atm and the half-life of HI at an initial pressure of 0.500 atm. Answer: The rate constant for the reaction is 0.0500 atm/s. The time it takes for the pressure to drop from 0.150 atm to 0.0432 atm is 2.14 seconds. The half-life of HI at an initial pressure of 0.500 atm is 5.00 seconds.

Step-by-step solution

(a) Rate constant calculation

To find the rate constant, we will use the zero-order kinetics formula for the given initial pressure (1.00 atm) and final pressure (0.200 atm) of HI over a time interval of 16.0 s. Plugging all the known values into the formula, we get: $$ 0.200\ \text{atm} = 1.00\ \text{atm} - k(16.0\ \mathrm{s}) $$ Next, we will solve for \(k\) by rearranging the equation: $$ k = \frac{1.00\ \text{atm} - 0.200\ \text{atm}}{16.0\ \mathrm{s}} = \frac{0.800\ \text{atm}}{16.0\ \mathrm{s}} $$ Finally, we will calculate the value of \(k\): $$ k = \frac{0.800\ \text{atm}}{16.0\ \mathrm{s}} = 0.0500\ \frac{\text{atm}}{\text{s}} $$

(b) Time calculation for pressure drop from 0.150 atm to 0.0432 atm

To find the time it takes for the pressure to drop from 0.150 atm to 0.0432 atm, we will use the rate constant calculated in (a) and the zero-order kinetics formula. The equation will now be: $$ 0.0432\ \text{atm} = 0.150\ \text{atm} - 0.0500\ \frac{\text{atm}}{\text{s}} t $$ Next, we will solve for \(t\) by rearranging the equation: $$ t = \frac{0.150\ \text{atm} - 0.0432\ \text{atm}}{0.0500\ \frac{\text{atm}}{\text{s}}} $$ Finally, we calculate the time it takes for the pressure to drop from 0.150 atm to 0.0432 atm: $$ t = \frac{0.1068\ \text{atm}}{0.0500\ \frac{\text{atm}}{\text{s}}} = 2.14\ \mathrm{s} $$

(c) Half-life calculation

The half-life (\(t_{1/2}\)) is the time required for the pressure of HI to drop by half. At an initial pressure of 0.500 atm, the final pressure will be: $$ 0.500\ \text{atm} \times 0.5 = 0.250\ \text{atm} $$ Now, we will use the rate constant calculated in (a) and the zero-order kinetics formula: $$ 0.250\ \text{atm} = 0.500\ \text{atm} - 0.0500\ \frac{\text{atm}}{\text{s}} t_{1/2} $$ We will rearrange the equation to solve for \(t_{1/2}\): $$ t_{1/2} = \frac{0.500\ \text{atm} - 0.250\ \text{atm}}{0.0500\ \frac{\text{atm}}{\text{s}}} $$ Lastly, we calculate the half-life: $$ t_{1/2} = \frac{0.250\ \text{atm}}{0.0500\ \frac{\text{atm}}{\text{s}}} = 5.00\ \mathrm{s} $$