\(20.0\,\mathrm{cm^3}\) of hydrogen peroxide decomposes to water and oxygen in the presence of a suitable catalyst.
\(160\,\mathrm{cm^3}\) of oxygen, measured at room conditions, is produced in \(5.00\,\mathrm{minutes}\).
What is the average rate of decomposition of hydrogen peroxide during this reaction period?
Correct Answer: \( \boxed{\mathrm{B}} \)
The reaction is:
\(\mathrm{2H_2O_2\rightarrow2H_2O+O_2}\)
Moles of oxygen produced:
\(\mathrm{n(O_2)=\dfrac{160}{24000}=6.67\times10^{-3}\,mol}\)
Moles of hydrogen peroxide decomposed:
\(\mathrm{n(H_2O_2)=2\times6.67\times10^{-3}=1.33\times10^{-2}\,mol}\)
Time \(=5.00\,\mathrm{min}=300\,\mathrm{s}\)
Average rate:
\(\mathrm{\dfrac{1.33\times10^{-2}}{300}=4.44\times10^{-5}\,mol\,s^{-1}}\)
Therefore, the correct answer is (B).
Aqueous hydrogen peroxide, \( \mathrm{H_2O_2} \), decomposes into water and oxygen in the presence of a suitable catalyst.
\(50\,\mathrm{cm^3}\) of a \(0.50\,\mathrm{mol\,dm^{-3}}\) solution of hydrogen peroxide produced \(120\,\mathrm{cm^3}\) of oxygen in \(2.0\) minutes.
The volume of gas was measured at room conditions.
What is the average rate of decomposition of hydrogen peroxide during this \(2.0\)-minute period?
Correct Answer: \( \boxed{\mathrm{C}} \)
The decomposition reaction is:
\( \mathrm{2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)} \)
At room conditions, \(1\,\mathrm{mol}\) of gas occupies \(24\,000\,\mathrm{cm^3}\).
Moles of \( \mathrm{O_2} \) produced:
\( \mathrm{\dfrac{120}{24000}=0.0050\,mol} \)
From the equation, moles of \( \mathrm{H_2O_2} \) decomposed:
\( \mathrm{2\times0.0050=0.010\,mol} \)
Volume of solution:
\( \mathrm{50\,cm^3=0.050\,dm^3} \)
Decrease in concentration of \( \mathrm{H_2O_2} \):
\( \mathrm{\dfrac{0.010}{0.050}=0.20\,mol\,dm^{-3}} \)
Time taken:
\( \mathrm{2.0\,min=120\,s} \)
Average rate of decomposition:
\( \mathrm{\dfrac{0.20}{120}=1.67\times10^{-3}\,mol\,dm^{-3}\,s^{-1}} \)
Therefore, the correct answer is (C).
The rate of the reaction between a reactive metal and an excess of a dilute acid is investigated.
The total volume of hydrogen gas produced is recorded every 30 seconds for 3 minutes.
| Time / s | Total volume of hydrogen gas / \( \mathrm{cm^3} \) |
|---|---|
| 0 | 0 |
| 30 | 64 |
| 60 | 105 |
| 90 | 132 |
| 120 | 151 |
| 150 | 161 |
| 180 | 167 |
The average rate of reaction during the first 30 seconds is \(P\).
The average rate of reaction during the last 30 seconds is \(Q\).
What is the value of \(P-Q\)?
Correct Answer: \( \boxed{\mathrm{B}} \)
Average rate during the first 30 s:
\(P=\dfrac{64-0}{30}=2.13\,\mathrm{cm^3\,s^{-1}}\)
Average rate during the last 30 s (150 s to 180 s):
\(Q=\dfrac{167-161}{30}=0.20\,\mathrm{cm^3\,s^{-1}}\)
Therefore,
\(P-Q=2.13-0.20=1.93\,\mathrm{cm^3\,s^{-1}}\)
Hence, the correct answer is (B).
