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1. The activation energy for the denaturation of a protein is 311.0 kJ/mol. At w

ID: 534654 • Letter: 1

Question

1. The activation energy for the denaturation of a protein is 311.0 kJ/mol. At what temperature will the rate of denaturation be 19 percent greater than its rate at 25.00°C?

_____°C

2. The thermal decomposition of N2O5 obeys first-order kinetics. At 45°C, a plot of ln[N2O5] versus t gives a slope of 7.30 × 104 min1. What is the half-life of the reaction?

___ × 10___min (Enter your answer in scientific notation.)

3. What rate constant can be determined from only the half-life of a reaction?

a) Zeroth

b) First

c) Second

d) Third

e) The rate constant cannot be determined using only a rate constant for any reaction.

4.

X ---> Y

X --> Z

5. Consider the following unbalanced equation:


A + B C


When C is being formed at the rate of 0.086 M/s, A is being consumed at a rate of 0.172 M/s and B is being consumed at a rate of 0.172 M/s. Balance the equation based on the relative rates of formation and consumption of products and reactants.

6. The reaction

2A + 3B C is first order with respect to A and B. When the initial concentrations are [A] = 1.80 × 102M and [B] = 2.11 ×103M, the rate is 4.89 ×104M/s.

Calculate the rate constant of the reaction.

Enter your answer in scientific notation.

___×10___

M

M1

s

s1

M/s

·

7. The rate constant for the following second-order reaction is 0.80/(M ·s) at 10°C.2NOBr(g) 2NO(g) + Br2(g)


Calculate the half-life when [NOBr]0 = 0.070 M.

_____ s

9.

rate = k[A]t

A compound X undergoes two simultaneous first-order reactions as follows:

X ---> Y

X --> Z



The ratio of k1/k2 at 40°C is 5.5. What is the ratio at 311°C? Assume that the frequency factors of the twoo reactions are the same.

Ratio = In [Alo kt

Explanation / Answer

1) Ans we know the  Arrhenius equation,

ln( k2/k1 ) = (Ea/R) x (1/T1 - 1/T2)

rate of denaturation k1 =100

rate of denaturation k2=100+19=119

Ea = 311.00 kJ/mol =311x1000 =311000 J/mol
R = 8.3144 ( is rate constant)
T1 = 25 °C =273+25 =298 K
T2 = ?

ln(119/100) = (311000/8.3144)x(1/298 - 1/T2)

ln(1.19) = (37404.98)(1/298 - 1/T2)

0.1740=(37404.98/298 )-(37404.98/T2)

(37404.98/T2)=125.5201 -0.1740

T2=37404.98/125.3461

T2=298.4 K

T2=298.4 - 273= 25.4°C

T2= 25.4°C