Below is the word equation for the reaction between sodium hydroxide and hydrochloric acid. Identify the reactant(s) in the reac
tion. Sodium hydroxide + Hydrochloric acid → Sodium chloride + Water
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]
10,573.4
Explanation:
Given parameters:
Dosage per patient = 73ap/lbs
Weight of patient = 65.7kg
Unknown:
Amount of grams/ap to be given to the patient;
Solution:
Since our final answer should be expressed in grams, let us convert all units into grams.
1000g = 1kg
Converting 65.7kg to g = 65.7 x 1000 = 65700g
73ap/lbs;
1lb = 453.6g
73ap/lbs x = 0.161ap/g
Now,
For a 65700g person, the number of grains app required;
65700 x 0.161 = 10,573.4grains ap per dose
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