For the reaction C_2H_6(g) rightarrow 2CH_3(g) rate - k[C_2H_6] If k = 5.50 Time

Question

For the reaction C_2H_6(g) rightarrow 2CH_3(g) rate - k[C_2H_6] If k = 5.50 Times 10^-4 s^-1 and [C_2H_6]_initial = 0.0200 M, calculate the rate of reaction after 1 hour. The rate of reaction is zero since the reaction is complete. 1.52 Times 10^-6 M s^-1 5.50 Times 10^-4 M s^-1 2.76 Times 10^-3 M s^-1 1.10 Times 10^-5 M s^-1 Given the reaction, aA + bB rightarrow^c dD + eE where C is the catalyst If we try , rate = k[A]^q[B]^r[C]^s for a generic rate law statement, which one of the following statements below is false? The exponenet q, r, and s are often integers. The exponent s must be determined experimentally. The exponents q and r are equal to the coefficients a and b, respectively. The overall order of the reaction is q + r + s. The symbol k represents the rate constant. The reaction: A + 3B rightarrow D + F was studied carefully and the following mechanism was finally determined. A + B doubleheadarrow C (fast) C + B doubleheadarrow D + E (slow) E + B doubleheadarrow F (very fast) The step with the largest activation energy is the first step the second step the third step none of the steps has an activation energy all of the steps have the same activation energy

Explanation / Answer

17) Given that rate = k[C2H6]

So, it is a first order reaction.

k = 5.5 x 10-4 s-1

[C2H6]initial = 0.02 M

For first order reaction, [At] = [Ao] e-kt

[At] = [C2H6] after 1 hr ( = 3600s )

[Ao]=  [C2H6]initial

[At] = [Ao] e-kt = (0.02 M ) e(-5.5 x 10-4 s-1 x 3600 s) = 2.76 x 10-3 M

Hence, [C2H6]after 1hr = 2.76 x 10-3 M

Then,

rate after 1 hr = k [C2H6]after 1hr =  5.5 x 10-4 s-1 x 2.76 x 10-3 M = 1.52 x 10-6 M.s-1

Therefore,

rate of reaction after 1 hr = 1.52 x 10-6 M.s-1

18) False statement = B

Reason:

The value of s my not be determined experimentally. It can be calculated stoichiometrically also.

19) Slowest step has largest activation energy.

Therefore, Answer = B ( 2nd step)

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