The decomposition of N_2O_4 is studied at 20 degreeC and at 80 degreeC. Which st

Question

The decomposition of N_2O_4 is studied at 20 degreeC and at 80 degreeC. Which statement explains why the rate at 80 degreeC is greater than at 20 degreeC? (A)The activation energy is lower at 80 degreeC. (B)The activation energy is higher at 80 degreeC. (C)The concentration of a gas increases with increasing temperature. (D)The number of molecules with enough energy to react is greater at 80 degreeC. 6.Compound A decomposes to form compound B. When 1/[A] is plotted versus time, a straight line with a slope of 0.53 M^-1 min^-1 results. What is the rate law? (A)rate = 0.53 M^-1 min^-1 (B)rate = 0.53 M^-1 min^-1 [A] (C)rate = 0.53 M^-1 min^-1[A]^2 (D)More information is necessary. 7.The rate law for the reaction A + B + C rigtarrow D + E is rate = K[A][B]^2 If the initial concentrations of A and C are doubled while all other conditions remain unchanged, by what factor will the reaction rate change? (A) 2 (B) 4 (C) 8 (D) The reaction rate will not change. 8.In this reaction, I^- is consumed at a rate of 2.5 M s^-What is the rate of formation of I_2 + 5I^- + 6H^+ rightarrow 3I_2 + 3H_2O (A) 0.25 (B) 1.5 (C) 2.5 (D) 4.2 9. For this two-step reaction sequence, identify the reaction intermediate and catalyst. step 1: H_2 + BR^- rightarrow BrO^- + H_2O step 2: H_2O_2 + BrO^- rightarrow Br^- + H_2O + O_2 Reaction intermediate catalyst (A) Br^- BrO^- (B) BrO^- H_2O (C) BrO^- Br^- (D) H_2O_2 Br^-

Explanation / Answer

5)

B) the activation energy is lower at 80 C

6) A --> B

given

1/[A] vs time is a straight line

so

it is a 2nd order reaction

1/[A] = kt + 1/[Ao]

slope = k = 0.53

now

rate = k [A]^2

rate = 0.53 [A]^2

the answer is C) rate = 0.53 M-1 min-1 [A]^2

7)

given

rate = k [A] [B]^2 [C]^0

rate 2 / rate 1 = ( 2A/A) ( B/B)^2 ( 2C / C)^0

rate 2 / rate 1 = 2

so

the rate is increased by a factor 2

so

the answer is option A) 2

8)

consider the given reaction

IO3- + 5I- + 6H+ ---> 3I2 + 3H20

we can see that

(-dI/dt) / 5 = (dI2/dt) / 3

2.5 / 5 = (dI2/dt) / 3

(dI2/dt) = 1.5

so

rate of formation of I2 is 1.5 M s-1

so

the answer is B) 1.5

9)

step 1 : H202 + Br- ---> BrO- + H20

step 2 : H202 + BrO- --> Br- + H20 + 02

we can see that

reaction intermediate : BrO-

catalyst : Br-

so

the answer is option B

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