Question

the reaction represented is first order with respect to \\(\ce{h2o2}\\).
\\\ce{2 h2o2 -> 2 h2o + o2}\\
an experiment is performed in which 100. ml of 0.20 \\(m\\) \\(\ce{h2o2}\\) is decomposed to \\(\ce{h2o}\\) and \\(\ce{o2}\\) at a temperature of 300 k. a chemist wants to triple the rate of the reaction.

which of the following experimental conditions would triple the rate of the reaction?

a) repeating the experiment with 50. ml of 0.60 \\(m\\) \\(\ce{h2o2}\\) at 300 k

b) repeating the experiment with 100. ml of 0.20 \\(m\\) \\(\ce{h2o2}\\) at 100 k

c) repeating the experiment with 200. ml of 0.067 \\(m\\) \\(\ce{h2o2}\\) at 300 k

d) repeating the experiment with 300. ml of 0.20 \\(m\\) \\(\ce{h2o2}\\) at 300 k

Explanation:

The reaction is first - order with respect to $\ce{H_{2}O_{2}}$, so the rate law is $rate = k[\ce{H_{2}O_{2}}]$, where $k$ is the rate constant (depends on temperature). To triple the rate, we need to triple the concentration of $\ce{H_{2}O_{2}}$ (since temperature should be constant to keep $k$ the same, or if temperature changes, it should not decrease the rate).

• Option A: Initial concentration $[\ce{H_{2}O_{2}}]_{initial}=0.20\ M$, new concentration $[\ce{H_{2}O_{2}}]_{new}=0.60\ M$. The ratio $\frac{0.60}{0.20} = 3$, so the rate will triple (temperature is same, $300\ K$).
• Option B: Lowering temperature to $100\ K$ will decrease the rate constant $k$, so the rate will not triple (and will likely be much lower).
• Option C: Calculate the new concentration effect. The moles of $\ce{H_{2}O_{2}}$ initially: $n = 0.20\ M\times0.100\ L=0.02\ mol$. New volume is $0.200\ L$, so new concentration $=\frac{0.02\ mol}{0.200\ L}=0.10\ M$ (wait, the option says $0.067\ M$? Wait, no, maybe miscalculation. Wait, if we want to keep moles same? Wait, no, the reaction is about rate which depends on concentration. Wait, initial moles: $0.20\ M\times0.1\ L = 0.02\ mol$. New volume $0.2\ L$, so concentration is $\frac{0.02}{0.2}=0.1\ M$, but the option has $0.067\ M$. Anyway, $0.067\ M$ is less than $0.20\ M$, so rate will decrease, not triple.
• Option D: Changing volume to $300\ mL$ with same concentration ($0.20\ M$) means moles increase, but concentration is same. So rate will not triple (rate depends on concentration, not volume directly when concentration is same).

Answer:

A. Repeating the experiment with 50. mL of 0.60 M $\ce{H_{2}O_{2}}$ at 300 K