NEED HELP ASAP! When heated, cyclopropane is converted to propene. Rate constants for this reaction at 470. °C and 510. °C are k1 = 1.10 × 10-4 s-1 and k2 = 1.02 × 10-3 s-1, respectively. Determine the activation energy, Ea, from these data.

1.06 × 10^5 J

Explanation:

We have to use the formula;

log(k2/k1) = Ea/2.303 ×R (1/T1 - 1/T2)

Given that;

k1 = 1.10 × 10-4 s-1

k2= 1.02 × 10-3 s-1

Ea=?

R=8.314 Jmol-1K-1

T1= 470. °C

T2= 510. °C

Hence;

log(1.02 × 10-3 s-1/1.10 × 10-4 =Ea/2.303 ×8.314(1/470 -1/510)

0.9672 = Ea/19.122(2.13 - 1.96) × 10^-3

0.9672= Ea/19.122(0.17) × 10^-3

0.9672= 1.7 ×10^-4 Ea/19.122

19.122 × 0.9672 = 1.74 × 10^-4Ea

Ea= 19.122 × 0.9672/1.74 × 10^-4

Ea= 18.495/1.74 × 10^-4

Ea= 1.06 × 10^5 J

the number of protons in the nucleus of a given element. has slight luster and conducts electricity at high temperatures like metal.but like a nonmetal,boron is brittle and does not conduct electricity well at low temperatures. in graphite, carbon is a soft,brittle solid.it is reflective,ductile,or malleable.