Question

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

rate = k[N2]^2 [H2]^2

Use this information to answer the questions below.

a. What is the reaction order in NO?
b. What is the reaction order in O2?
c. What is overall reaction order?
d. At a certain concentration of NO and O2, the initial rate of reaction is 6.0 x 10^4 M / s. What would the initial rate of the reaction be if the concentration of NO were halved?
e. The rate of the reaction is measured to be 79.0 M / s when [NO] = 1.2 M and [O2] = 1.9 M. Calculate the value of the rate constant. Round your answer to 2 significant digits.

Answer 1

Answer:

a) The reaction order in NO is 2

b) The reaction order in O₂ is 2

c) The overall order of the reaction is 4

d) If NO's concentration is halved, the reaction rate is quartered and is equal to 1.5 × 10⁴ M/s.

e) Reaction rate constant, k = 15.197 s/M³ = 15 s/M³ to 2s.f

Explanation:

The reaction order of a reaction with respect to a reactant is the power the reactant carries in the rate law.

a) The reaction order in NO is 2.

b) The reaction order in O₂ is 2.

c) Overall order is the sum of all the powers in the rate law, and for this reaction, it is 2 + 2 = 4

d) If the other parameters are constant, the rate of the reaction is directly proportional to the square of the concentration of NO

r = K [NO]²

At point 1, let r = r₁ and [NO] = [(NO)₁]

At point 2, let r = r₂ and [NO] = [(NO)₂]

r₁ = K [(NO)₁]²

At point 2,

r₂ = K [(NO)₂]²

But [(NO)₂] = [(NO)₁/2]

So, r₂ = K [(NO)₁/2]² = K [(NO)₁]²/4

And r₁ = K [(NO)₁]²,

So, r₂ = r₁/4

Meaning the reaction rate becomes (6.0 × 10⁴)/4 = 1.5 × 10⁴ M/s

e) The rate of the reaction is measured to be 79.0 M / s when [NO] = 1.2 M and [O2] = 1.9 M. Calculate the value of the rate constant.

r = k [NO]² [O₂]²

79 = k × 1.2² × 1.9²

k = 15.197 s/M³

Answer 2

a. The reaction order in nitrogen oxide (NO) is 2.

b. The reaction order in oxygen gas ($O_2$) is 2.

c. The overall reaction order is equal to 4.

d. The initial rate of the reaction if the concentration of NO were halved is $1.5 \times 10^4\;M/s$.

e. The value of the rate constant to 2 significant digits is $15.20\;s/M^3$

Given the following data:

• Rate law = $k[NO]^2[H_2]^2$

• Initial rate = $6.0 \times 10^4\;M/s$

• Rate = 79.0 M

• Concentration of [NO] = 1.2 M

• Concentration of [$O_2$] = 1.9 M.

What is the rate law?

Rate law refers to a chemical equation that is mainly used to relate the initial (forward) reaction rate with respect to the concentrations or pressures of the chemical reactants and constant parameters.

In Chemistry, the rate law is given by this formula:

$R = k[A]^x[B]^y$

Where:

• k is the rate constant.

• A is the concentration of reactant A.

• B is the concentration of reactant B.

• x is the reaction order with respect to reactant A.

• y is the reaction order with respect to reactant B.

a. The reaction order in nitrogen oxide (NO) is 2.

b. The reaction order in oxygen gas ($O_2$) is 2.

c. To calculate the overall reaction order:

Overall order is calculated by adding or summing up all the reaction order (powers) in the rate law.

$Overall \;order = 2 + 2$

Overall order = 4

d. To calculate the initial rate of the reaction if the concentration of NO were halved:

Note: The rate of a chemical reaction is directly proportional to the square of the concentration of nitrogen oxide (NO) when the other parameters are kept constant.

Mathematically, this is given by the expression:

$R_2=\frac{R_1}{4} \\ \\ R_2=\frac{6.0 \times 10^4}{4} \\ \\ R_2 = 1.5 \times 10^4\;M/s$

e. To calculate the value of the rate constant to 2 significant digits:

$R= k[NO]^2[H_2]^2\\ \\ 79.0=k\times 1.2^2 \times 1.9^2\\ \\ 79.0=5.1984k\\ \\ k=\frac{79.0}{5.1984} \\ \\ k = 15.20\;s/M^3$

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