Given the following rate law, how does the rate of reaction change if the concen

Question

Given the following rate law, how does the rate of reaction change if the concentration of Y is doubled? Rate = K [ X ]^2 [ Y ]^3
a)The rate of the rxn will increase by factor of 2. b) “. “. “. “. “. “. “. “. 4. c). “. “. “. “. “. “. “. “. 8. d). “. “. “. “. “. “. “. “. 9. Given the following rate law, how does the rate of reaction change if the concentration of Y is doubled? Rate = K [ X ]^2 [ Y ]^3
a)The rate of the rxn will increase by factor of 2. b) “. “. “. “. “. “. “. “. 4. c). “. “. “. “. “. “. “. “. 8. d). “. “. “. “. “. “. “. “. 9. Rate = K [ X ]^2 [ Y ]^3
a)The rate of the rxn will increase by factor of 2. b) “. “. “. “. “. “. “. “. 4. c). “. “. “. “. “. “. “. “. 8. d). “. “. “. “. “. “. “. “. 9.

Explanation / Answer

Given Rate law is:
rate = k[X]^2[Y]^3
rate new / rate old = ([X]new/[X]old)^2*([Y]new/[Y]old)^3
here:
[X]new/[X]old = 1
[Y]new/[Y]old = 2
putting values
rate new / rate old = (1.0)^2*(2.0)^3
rate new / rate old = 8
Answer: c

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.