Solution X is a strong base with a pH of 12. Solution X is mixed with solution Y,and the pH of the resulting mixture is 8.Based
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Answer:
- Question 1: <u>Both the rate of reaction of H₂O₂ and the rate of formation of H₂O are:</u>
- Question 2: <u>0.60 moles of O₂ are formed in the first 50 s of reaction.</u>
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Explanation:
The coefficients of the balanced chemical equation tells the relation of formation of the reactants into the products, which is also the relation between their rates of reaction.
The balanced chemical equation (given) is:
Then, in the same time that 2 moles of H₂O₂ react, 2 moles of H₂O and 1 mol of O₂ are formed.
<u>1. Question 1.</u>
That means that the rate of formation of O₂ is half the rate of reaction of H₂O₂ and half the rate of formation of H₂O.
Hence, if the instantaneous rate of formation of O₂ is 3.3×10⁻³moles/(liters×seconds), then the rate of reaction of H₂O₂ and the rate of formation of H₂O is twice:
<u>2. Question 2.</u>
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First you must find how much the concentration has changed.
You can read the initial concentration of H₂O₂ on the graph. It is the y-intercept. It is 1.0M.
You can read the approixmate concentration of H₂O₂ at the time 50 s. It is about 0.2M.
Then the change in concentration in the first 50 s of reaction is about 1.0M - 0.2M = 0.8M.
Now you can convert the concentration 0.8M into number of moles, using the volume (1.5 liters), assuming the liquid volume does not change, and the molarity definition (molarity is the number of moles of solute in one liter of solution)
Hence, 1.2 mol of H₂O₂ have reacted in the first 50s of reaction.
From the coefficients of the chemical equation, the number of moles of O₂ formed is half the number of moles of H₂O₂ that react.
Thus, 1.2mol/2 = 0.60 moles of O₂ are formed in the first 50 s of reaction.