What is the talc law for the reaction? (A) Rate = k[NO][O_2] (B) Rate = k[NO][O_

Question

What is the talc law for the reaction? (A) Rate = k[NO][O_2] (B) Rate = k[NO][O_2] (C) Rate = k[NO][O_2]^2 (D) Rate = k[NO]^2[O_2] For a particular reaction, the rate constant doubles when the temperature is raised from 300 K to 310 K. What is the activation energy (in kJ mol^-1)? (A) -0.0596 kJ middot mol^-1 (B) 0.l55 kJ middot mol^-1 (C) 53.6 kJ middot mol^-1 (D) 155 kJ mol^-1 For all exothermic reactions, the activation energy for the reverse reaction is the activation energy of the forward reaction. (A) smaller than (B) the same as (C) greater than (D) It cannot be determined from the information given. Why does the rate of reaction increase as the temperature increases? (A) The rate constant increases. (B) The rate constant decreases (C) Activation energy increases (D) Activation energy decreases.

Explanation / Answer

Q1.

From experiment 1 and 2

NO is kept constant; now notice that O2 is doubled

the rate is also doubled, therefore

ratio must be first order with respect to O2

now.. from expeirment 1 and 3 (we use this since O2 is now constant)

this implies

as we double the NO concentration from 0.045 to 0.090; we double the rate from 2*10^-5 to 4*10^-5

therefore, it is also first order with respect to NO

the overall reaction

Rate = K*[NO][O2]

choose B

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