Almost done with my homework but can\'t seem to work this problemout either. Any

Question

Almost done with my homework but can't seem to work this problemout either. Any help?
The first-order rate constant for the reaction of methylchloride (CH3Cl) with water to produce methanol (CH3OH) andhydrochloric acid (HCl) is 3.32 multiplied by 10-10s-1 at 25°C.Calculate the rate constant at 47°C if the activation energy is116 kJ/mol. (answer in s^-1 units)
The first-order rate constant for the reaction of methylchloride (CH3Cl) with water to produce methanol (CH3OH) andhydrochloric acid (HCl) is 3.32 multiplied by 10-10s-1 at 25°C.Calculate the rate constant at 47°C if the activation energy is116 kJ/mol. (answer in s^-1 units)

Explanation / Answer

Equation: ln(k1/k2) = (Ea/R)(1/T2 - 1/T1) k1 = 3.32e-10 k2 = unknown Ea = 116000 J/mol R = 8.3145 T2 = 47 + 273 = 320 K T1 = 25 + 273 = 298 K ln(3.32e-10/x) = (116000/8.3145)(1/320 - 1/298) x = 8.3e-9 s-1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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