Question

Given the following equation, C2H₅I (g) ⇒ C2H4 (g) + HI (g) k= 4.0 * 10^-3 s^ -1 at 700K

a. Write the rate law.

b. what is the rate of reaction when {C₂H₅I}=3.3x10^-4mol/L?

c. if the concentration is doubled, what is the effect on the reaction?

d. What is the half life?

2. Given the following equation 2C₂H₂=C₄H₄ K=0.178 L/mol at 770K

a. write the rate law

b. what is the rate of reaction when C₂H₂=0.195M

c. what is the half life

d. How much is required for 70% of the C₂H₂ to be used up?

Answer 1

1.

(a)

Rate law,

Rate = k[C2H5I]

From the units of k , it is understood that the reaction is first order reaction.

(b)

Rate = 4.0 * 10-3 * 3.3 * 10-4

Rate = 1.32 * 10-6 M/s

(c)

If the concentration is doubled the rate of reaction is also doubled.

(d)

Half life is the time required for the conversion of half of the reactant into product.

Half life time = 0.693 / k = 0.693 / (4.0 * 10-3) = 173. s

2.

(a)

From the units of k , it is understood that the reaction is second order reaction.

Rate = k[C₂H₂]²

(b)

Rate = 0.178 * (0.195)² = 6.77 * 10^-3 M/s

(c)

Half life time = 1 / k[A]0 = 1 / (0.178 * 0.195) = 28.8 s

(d)

Initial [A]0 = 0.195 M

Final [A] = 0.195 - (0.195 * 70 / 100) = 0.0585 M

Second order rate constant equation is,

1 / [A] = kt + 1 / [A]₀

1 / 0.0585 = (0.178 * t) + ( 1 / 0.195)

t = 67.2 s