The activation energy of a certain reaction is 37.2 kj/mol . at 20 ∘c, the rate constant is 0.0130 s−1. at what temperature would this reaction go twice as fast? express your answer numerically in degrees celsius

At 34°c

Explanation:

Using The Arrhenius Equation:

k = Ae − Ea/RT

k represents rate constant

A represents frequency factor and is constant

R represents gas constant which is = 8.31J/K/mol

Ea represents the activation energy

T represents the absolute temperature.

By taking the natural log of both sides,

ln k = ln A- Ea/RT

Reactions at temperatures T1 and T2 can be written as;

ln k1= ln A− Ea/RT1

ln k2= ln A− Ea/RT2

Therefore,

ln(k1/k2) = −Ea/RT1 + Ea/RT2

Since k2=2k1 this becomes:

ln(1/2) = Ea/R*[1/T2 − 1/T1]

Theefore,

-0.693 = 37.2 x 10^3/8.31 * [ 1/T2 - 1/293]

1/T2 - 1/293 = -1.55 x 10^-4

1/T2 = -1.55 x 10^-4 + 34.13x 10^-4

1/T2 = 32.58 x 10^-4

Therefore T2 = 307K

T2 = 307 - 273 = 34 °c

pecialized keyboards

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.