In a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH2. In several experiments the rate constant k was determined at different temperatures. A plot of ln(k) versus 1/T was constructed that resulted in a straight line with a slope of 21.10 3 104 K and a y intercept of 33.5. Assume that k has units of s21. a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor A. c. Calculate the value of k at 258C.

Ea= -175.45J

A= 3.5×10^14

k=3.64 ×10^14 s^2.

Explanation:

From

ln k= -(Ea/R) (1/T) + ln A

This is similar to the equation of a straight line:

y= mx + c

Where m= -(Ea/R)

c= ln A

y= ln k

a)

Therefore

21.10 3 104= -(Ea/8.314)

Ea=-( 21.10 3 104×8.314)

Ea= -175.45J

b) ln A= 33.5

A= e^33.5

A= 3.5×10^14

c)

k= Ae^-Ea/RT

k= 3.5×10^14 × e^ -(-175.45/8.314×531)

k = 3.64 ×10^14 s^2.

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.