During this reaction: [tex]P_4 + 5 O_2 \rightarrow P_4O_{10}[/tex], 1.5 moles of product were made in 30 seconds. What is the rate of reaction?

A. [tex]380 \, \text{g/min}[/tex]
B. [tex]0.011 \, \text{g/min}[/tex]
C. [tex]850 \, \text{g/min}[/tex]
D. [tex]210 \, \text{g/min}[/tex]



Answer :

To determine the rate of the reaction given the formation of \( P_4O_{10} \), let's follow the steps carefully:

1. Given Data:
- Moles of \( P_4O_{10} \) formed: 1.5 moles
- Time taken: 30 seconds

2. Convert time to minutes:
- Since reaction rates are given in grams per minute, we need to convert the time from seconds to minutes.
[tex]\[ \text{Time (minutes)} = \frac{\text{Time (seconds)}}{60} = \frac{30}{60} = 0.5 \text{ minutes} \][/tex]

3. Calculate the rate in moles/minute:
- The rate of reaction in terms of moles per minute is computed as:
[tex]\[ \text{Rate (moles/minute)} = \frac{\text{Moles of } P_4O_{10}}{\text{Time (minutes)}} = \frac{1.5 \text{ moles}}{0.5 \text{ minutes}} = 3 \text{ moles/minute} \][/tex]

4. Determine the molar mass of \( P_4O_{10} \):
- To convert the moles per minute to grams per minute, we need the molar mass of \( P_4O_{10} \).
- The atomic mass of Phosphorus (P) is approximately 30.974 g/mol.
- The atomic mass of Oxygen (O) is approximately 15.999 g/mol.

Thus, the molar mass of \( P_4O_{10} \) can be calculated as:
[tex]\[ \text{Molar mass of } P_4O_{10} = 4 \times 30.974 + 10 \times 15.999 = 123.896 + 159.99 = 283.886 \text{ g/mol} \][/tex]

5. Calculate the rate in grams per minute:
- Using the rate in moles per minute and the molar mass, we find the rate in grams per minute:
[tex]\[ \text{Rate (grams/minute)} = \text{Rate (moles/minute)} \times \text{Molar mass (g/mol)} = 3 \text{ moles/minute} \times 283.886 \text{ g/mol} = 851.658 \text{ g/minute} \][/tex]

6. Compare with Given Options:
- The calculated rate in grams per minute is 851.658 g/min. We need to compare this with the provided options: \( 380 \text{ g/min}, 0.011 \text{ g/min}, 850 \text{ g/min}, 210 \text{ g/min} \).
- The closest value to 851.658 g/min from the provided options is \( 850 \text{ g/min} \).

Therefore, the rate of reaction is [tex]\( \boxed{850 \text{ g/min}} \)[/tex].

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