# The rate of a certain reaction

The rate of a certain reaction is given by the following rate law: rate=kNO2O2

Use this information to answer the questions below.

 What is the reaction order in NO? What is the reaction order in O2? What is overall reaction order? At a certain concentration of NO and O2, the initial rate of reaction is 79.0 M / s. What would the initial rate of the reaction be if the concentration of NO were doubled? Round your answer to 3 significant digits. Ms The rate of the reaction is measured to be 6.0 x 104 M / s when [NO] = 0.70 M and [O2] = 1.1 M. Calculate the value of the rate constant. Round your answer to 2 significant digits. =k⋅M−2s−1

# The rate of a certain reaction

The rate of a certain reaction is given by the following rate law:

rate =kNO2O2

Use this information to answer the questions below.

 What is the reaction order in NO? What is the reaction order in O2? What is overall reaction order? At a certain concentration of NO and O2, the initial rate of reaction is 79.0 M / s. What would the initial rate of the reaction be if the concentration of NO were doubled? Round your answer to 3 significant digits. Ms The rate of the reaction is measured to be 6.0 x 104 M / s when [NO] = 0.70 M and [O2] = 1.1 M. Calculate the value of the rate constant. Round your answer to 2 significant digits. =k⋅M−2s−1

# The rate of a certain reaction

Question
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The rate of a certain reaction was studied at various temperatures. The table shows temperature (?)(T) and rate constant (?)(k) data collected during the experiments. Plot the data to answer the questions.

What is the value of the activation energy, ?aEa , for this reaction?

? (?)T (K) ? (?−1)k (s−1)
400400 0.0001080.000108
420420 0.0006260.000626
440440 0.003100.00310
460460 0.01340.0134
480480 0.05110.0511
500500 0.1750.175
520520 0.5470.547
540540 1.571.57
560560 4.184.18
580580 10.410.4

?a=Ea=

kJ⋅mol−1kJ⋅mol−1

What is the value of the pre‑exponential factor (sometimes called the frequency factor), ?A , for this reaction?
?=A=

s−1
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Step 1

Given

Table for temperature and rate constant values

T (K) k (s−1)
400 0.000108
420 0.000626
440 0.00310
460 0.0134
480 0.0511
500 0.175
520 0.547
540 1.57
560 4.18
580 10.4

To find-

1. Value of activation energy – Ea
2. Value of pre‑exponential factor A
Step 2

Two Forms of Arrhenius equation-

K= A e−EaRTe-EaRT

lnK = lnA – EaRTEaRT

were,

K = rate constant

A = pre‑exponential factor

Ea = activation energy

R = gas constant

T = temperature in K

Step 3

Approach

lnK = lnA – EaRTEaRT

This equation is in the form of-

y= mx + C

therefore,

y = lnk

x = 1T1T

C = lnA is intercept

m =- EaREaR is slope

• Graph can be plot by taking lnK values on x axis and 1T1T value on Y axis.
• slope =  -EaREaR ; value of Ea can be found out form slope

Step 4
• Calculating values of lnK and 1T1T
T  (K) 1T1T k (s−1) lnK
400 0.0025 0.000108 -9.133
420 0.00238 0.000626 -7.376
440 0.00227 0.00310 -5.776
460 0.00217 0.0134 -4.312
480 0.00208 0.0511 -2.974
500 0.002 0.175 -1.743
520 0.00192 0.547 -0.603
540 0.00185 1.57 -1.852
560 0.00179 4.18 1.43
580 0.00172 10.4 2.342

Step 5

Step 6

Equation form graph is

y= –13969 x + 25.889

Comparing with y= mx  + c

therefore,

slope = m = -13969 =- EaREaR

C = 25.889 = lnA

-13969 =– Ea8.314 J  mol−1 Ea8.314 J  mol-1

Ea = 116138 J mol-1

Ea = 116138 Jmol Jmol  x 1 KJ1000 J1 KJ1000 J = 116 KJmol KJmol

Step 7

C = 25.889 = lnA

A = anti ln 25.889

A = 1.75 x 1011 s-1

Step 8

1. Value of activation energy (Ea) = 116 KJmol KJmol
2. Value of pre‑exponential factor A = 1.75 x 1011 s-1

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