Question

The reaction 2a + 3b → c is first order with respect to a and

b. when the initial concentrations are [a] = 1.60 × 10−2 m and [b] = 2.67 × 10−3 m, the rate is 2.65 × 10−4 m/s. calculate the rate constant of the reaction. enter your answer in scientific notatio

Answer 1

The rate constant of the reaction 2a + 3b → c is calculated to be 6.22 × 10⁻⁲ m²/s using the given initial concentrations and the reaction rate, applying the first order rate law formula.

Step-by-step explanation:

To calculate the rate constant for the first order reaction with respect to both a and b (2a + 3b → c), we need to use the rate law formula which for this reaction is given as:

rate = k[a][b]

Where rate is the reaction rate, k is the rate constant, and [a] and [b] are the concentrations of reactants a and b, respectively. Given the initial concentrations and the reaction rate, the rate constant can be determined by rearranging and solving the equation:

2.65 × 10⁻⁴ m/s = k(1.60 × 10⁻⁲ m)(2.67 × 10⁻⁳ m)

Solving for k, we get:

k = (2.65 × 10⁻⁴ m/s) / ((1.60 × 10⁻⁲ m) (2.67 × 10⁻⁳ m))

After performing the calculations, we find:

k = 6.22 × 10⁻⁲ m²/s

Thus, the rate constant of the reaction, in scientific notation, is 6.22 × 10⁻⁲ m²/s.

Answer 2

6.30 M⁻¹s⁻¹.

Step-by-step explanation:

As the reaction is first order with respect to a and b.

The rate of the reaction = k [a][b],

where, k is the rate constant of the reaction.

The rate of the reaction = 2.65 × 10⁻⁴ M/s,

[a] = 1.60 × 10⁻² M,

[b] = 2.67 × 10⁻³ M.

∴ k = (the rate of the reaction / [a][b]) = (2.65 × 10⁻⁴ M/s) / (1.60 × 10⁻² M)( 2.67 × 10⁻³ M) = 6.30 M⁻¹s⁻¹.