Chemistry, 22.02.2020 18:57 kaleahearly123

Consider the following reaction:

2H2O2(aq)→2H2O(l)+O2(g)

The graph (Figure 1) shows the concentration of H2O2 as a function of time.
If the instantaneous rate of formation of O2 is 3.3(10^-3) moles/(litersseconds), then...
If the initial volume of the H2O2 solution is 1.5 L , what total amount of O2 (in moles) is formed in the first 50 s of reaction?
Express your answer using two significant figures.

Consider the following reaction: 2H2O2(aq)→2H2O(l)+O2(g) The graph (Figure 1) shows the concentratio

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Answer from: knj281

The coefficients of the balanced chemical equation tell the relation of formation of the reactants into the products, which is also the relation between their rates of reaction.

In the given reaction, 2 moles of H₂O₂ react, 2 moles of H₂O, and 1 mol of O₂ are formed.

In question 1

, 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.

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: $2*3.3*10^{-3}$

After solving it, the value is $6.6 * 10^{-6}$

In question 2. we must find how much the concentration has changed.

The initial concentration of H₂O₂ is on the graph. It is the y-intercept. It is 1.0M. After approximate the 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, convert the concentration 0.8M into the 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)

$0.8 = \frac{mole}{volume}$

After, putting the value,

moles = 0.8 * 1.5 = 1.2mol

Hence, 1.2 mol of H₂O₂ has reacted in the first 50s of reaction.

Thus, $\frac{1.2}{2} = 0.60 moles$ O₂ is formed in the first 50 s of reaction.

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Answer from: DragonLovely

Question 1: Both the rate of reaction of H₂O₂ and the rate of formation of H₂O are:

$6.6\cdot 10^{-3}mol/(liter.s)$

Question 2: 0.60 moles of O₂ are formed in the first 50 s of reaction.

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:

$2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)$

Then, in the same time that 2 moles of H₂O₂ react, 2 moles of H₂O and 1 mol of O₂ are formed.

1. Question 1.

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:

$2\times 3.3\cdot 10^{-3}mol/(liter.s)=6.6\cdot 10^{-3}mol/(liter.s)$

2. Question 2.

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)

$\Delta [H_2O_2]=\Delta moles/Volume(liters)$

$0.8M=\Delta moles/1.5liter\\\\\Delta moles=0.8M\times 1.5liter=1.2mol$

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.

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Consider the following reaction:

2H2O2(aq)→2H2O(l)+O2(g)

The graph (Figure 1...