I have 2 rate order questions The thermal decomposition of acetaldehyde (CH 3 CH

Question

I have 2 rate order questions

The thermal decomposition of acetaldehyde (CH3CHO) is asecond order reaction. At 700.0 K, the rate constant is 0.011L/mol·s. At 760.0 K, the rate constant increases to0.105 L/mol·s. What is the activation energy requiredfor this reaction?

1. 1.66 x 10^5kJ/mol
2. -1.66 x 10^5kJ/mol
3. 1.66 x 10^2kJ/mol
4. -1.66 x 10^2 kJ/mol


AND

A reaction has an activation energy of 123 kJ/mol and a rateconstant of 0.200 s-1 at 311 K. At whattemperature will the rate constant be double that at 311K?

1. 622 K
2. 316 K
3. 156 K
4. -23 K

Any help is appreciated, even just the formulas needed.

Explanation / Answer

k = A*exp(-E/RT) 0.011 = A*exp(-E/700R) 0.105 = A*exp(-E760R) Divide both equations 0.011/0.105 = exp(-E(1/700R-1/760R) ln(0.011/0.105) = -E(1/700R - 1/760R) R = 8.315 J/mol/K -2.256 = -E(1.356e-5) E = 166 kJ/mol Pick choice 3 For part 2, k1 = A*exp(-E/RT1) k2 = A*exp(-E/RT2) k1/k2 = 0.5 (k2 = 2*k1 = 2*0.200) E = 123 kJ/mol = 123000 J/mol k1/k2 = exp(-E/R *(1/T1-1/T2)) ln(k1/k2) = ln(0.5) = -(E/R)*(1/T1-1/T2) 1/T2 = 1/T1 + (R/E)*ln(0.5) 1/T2 = 1/311K + (8.314/123000)*ln(0.5) = 0.0031686 T2 = 315.6 K

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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