If Ea for a certain biological reaction is 50.0 kJ/mol, bywhat factor (how many

Question

If Ea for a certain biological reaction is 50.0 kJ/mol, bywhat factor (how many times will the rate of this reaction inrease)when body temperature increases from 37 degrees to 40degrees. ln(k)=Ea/R(1/T1-1/T2) ln(k)=50/8.314(1/310-1/313) ln(k)=1.86x10^-4 So to find the ln of k do I take e^1.86x10^-4? or how do Isolve it from here? If Ea for a certain biological reaction is 50.0 kJ/mol, bywhat factor (how many times will the rate of this reaction inrease)when body temperature increases from 37 degrees to 40degrees. ln(k)=Ea/R(1/T1-1/T2) ln(k)=50/8.314(1/310-1/313) ln(k)=1.86x10^-4 So to find the ln of k do I take e^1.86x10^-4? or how do Isolve it from here?

Explanation / Answer

Formula:                ln (k2 / k1) = Ea / R ( 1/T1 - 1/T2) Data:               T1 = 273 + 37 K                      =310K               T2 = 273 + 40 K                      =313 K                Ea = 50.0 kJ/mol                R = 8.314 J / mol .K   ln(k2 / k1)  = 50 * 1000 J / mol /8.314 J / mol.K ( 1 / 310 K - 1 / 313 K)                    = 0.185       (k2 / k1) = 1.2             k2   = 1.2 k1 By a factor of 1.2 rate of the reaction increases onincreasing the temperature from 37 degrees to 40 degrees.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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