A) What mass in grams of H20 is needed to react completely with 40.0 g of
1 answer:
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In general, we have this rate law express.:
![\mathrm{Rate} = k \cdot [A]^x [B]^y](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5Ey)
we need to find x and y
ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).
then we go to compare two experiments in which only one concentration is changed
compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]
Δ[B]= 0.3 / 0.1 = 3
now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:
ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...
solve for y in the equation![\Delta \mathrm{Rate} = \Delta [B]^y](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BB%5D%5Ey)
To this point,![\mathrm{Rate} = k \cdot [A]^x [B]^1](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5E1%20)
do the same to find x.
choose two experiments in which only the concentration of B is unchanged:
Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4
solve for x for![\Delta \mathrm{Rate} = \Delta [A]^x](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BA%5D%5Ex)
the rate law is
Rate = k·[A]²[B]
we need to find x and y
ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).
then we go to compare two experiments in which only one concentration is changed
compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]
Δ[B]= 0.3 / 0.1 = 3
now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:
ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...
solve for y in the equation
To this point,
do the same to find x.
choose two experiments in which only the concentration of B is unchanged:
Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4
solve for x for
the rate law is
Rate = k·[A]²[B]
Answer:The correct answer is ;
The oxidation state of nitrogen in NO changes from +2 to 0, and the oxidation state of carbon in CO changes from +2 to +4 as the reaction proceeds.
Explanation:
In an oxidation recation addition of oxygen atom takes place or loss of electrons takes place.
In an reduction reaction removal of oxygen atom takes place or gain of electrons takes place.
In the given reaction , the nitrogen atom is present in +2 oxidation state in NO molecule and present in 0 oxidation state in molecule. Hence, nitrogen is getting reduced that is reduction reaction. NO is oxidizing agent
In the given reaction , the carbon atom is present in +2 oxidation state in CO molecule and present in +4 oxidation state in molecule. Hence ,carbon is getting oxidized that is oxidation reaction. CO is a reducing agent.
Answer : The rate of consumption of oxygen = 0.245 mol/s
Solution : Given,
Rate at which Hydrogen burns = 0.49 mol/s
The Reaction is,
In this reaction, 2 moles of hydrogen react with the 1 mole of oxygen.
The rate at which oxygen burns is equal to the half of rate at which hydrogen burns.
Rate at which Oxygen burns = × 0.49 mol/s
= 0.245 mol/s