Still angsting for more? Try this Extra Credit question. (after you have tried t

Question

Still angsting for more? Try this Extra Credit question. (after you have tried the first 2 problems).. You're welcome H_2(g) + I_2(g) reversiblearrow2 HI(g) The table below gives data for a reaction rate study of the reaction represented above. Order of reaction with respect to H_2 is 1. For example. I + have 2 times of the initial concentration of H_2 then second one to first one, Also The Ic of Cl_2 constant result in 2 times of the section rate. Determine the order of the reaction with respect to H_2 and justify your answer. Determine the order of the reaction with respect to I_2 and justify your answer. Write the overall rate law for the reaction. Calculate the value of the rate constant and show the units. Predict the initial rate of the reaction if the initial concentration of H_2 is 0.00300 mol L-1 and the initial concentration of I_2 is 0.000500

Explanation / Answer

Let us take the rate expression as rate = k [H2]x [I2]y

a) To find order with respect to H2

compare rate for experiment 1 and 2

Rate 1 /rate 2= 1.82x10-12 /3.64x10-12

= k [0.001 ]x [0.0005]y / k [0.002 ]x [0.0005]y

thus 1/2 = 1/2x   and thus x = 1

hence order with respect to H2 is 1.

b) similarly order with respect to I2 is calculated by comparing rate 2 and 3.

  

Rate 2 /rate 3 = 3.64x10-12 /1.82x10-12  

= k [0.002 ][0.0005]y / k [0.002 ][0.00025]y

thus 2 = 2y   and y =1

Thus order with respect to I2 is = 1

c) The rate law is   

Rate = k [H2] [I2]

d) k = rate /[H2][I2] substituting the values form experiment 1

k = 1.82x10-12 M/s/(0.001) M x (0.0005)M

= 1.82 x10-5 /M.s

= 1.82 x10-5 Lmol-1s-1

e)When [H2] = 0.003M , the rate from experiment 1 will be tripled

Thus initial rate of reaction = 5.46 x10-12M/s

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