Iodine, I_2, is a blue-black solid, but it easily vaporizes to give a violet vap

Question

Iodine, I_2, is a blue-black solid, but it easily vaporizes to give a violet vapor. At high temperatures, this molecular substance dissociates to atoms: I_2(g) 2I_(g) An absent-minded professor measured equilibrium constants for this dissociation at two temperatures, 700 degree C and 800 degree C. He obtained the K_p values 0.001106 and 0.001745, but couldn't remember w u t value went with what temperature. Based on the equation, determine which temperature assigned to each equilibrium constants, and explain your reasoning.

Explanation / Answer

According to Arrhenius Equation , K = A e -Ea / RT

Where

K = rate constant / Equilibrium constant

T = temperature

R = gas constant = 8.314 J/mol-K

Ea = activation energy

A = Frequency factor

So from this equation as T increases the valave of K also increases.

As we increasing the temperature the number of collisions between the reacting molecules increases there by the formation of products increases so the rate of the reaction will increases.

Given the values of Kp as 0.001106 & 0.001745

The temperatures are 700 oC & 800 oC

So more value of Kp at more temperature

Therefore the value of Kp at 700 oC is 0.001106

              the value of Kp at 800 oC is 0.001745

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