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The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.30 kJ/mol and a frequency factor of A = 6.00×108M−1⋅s−1 . The reaction is believed to be bimolecular:NO(g)+F2(g)→NOF(g)+F(g)What is the rate constant at 695 ∘C ?Express your answer to three significant digits with the appropriate units. For compound units, place a multiplication dot between units (e.g. J⋅mol−1⋅K−1).(I got 2.74*10^8 M-1s-1 but it is wrong. I got a message saying "Enter your answer using dimensions of rate constant units for second-order reaction". Can someone please help me?)

Respuesta :

dsdrajlin

Answer:

2.74 × 10⁸ M⁻¹.s⁻¹ = 2.74 × 10⁸ L.mol⁻¹.s⁻¹

Explanation:

Let's consider the following second-order reaction.

NO(g) + F₂(g) → NOF(g) + F(g)

We can find the rate constant (k) using the Arrhenius equation.

[tex]k=A \times e^{-Ea/R\times T}[/tex]

where,

  • A: frequency factor
  • Ea: activation energy
  • R: ideal gas constant
  • T: absolute temperature (695°C + 273.15 = 968 K)

[tex]k=6.00 \times 10^{8} M^{-1}.s^{-1} \times e^{-6.30\times 10^{3}J /(8.314 J/mol.K)\times 968 K} = 2.74 \times 10^{8} M^{-1}.s^{-1}[/tex]

The dimensions of the rate constant units would be:

[tex]2.74[/tex] × [tex]10^8 M^{-1}[/tex][tex].S^{-1}.[/tex] = [tex]2.74[/tex] × [tex]10^8 L.mol^{-1} .s^{-1}[/tex]

Rate Constant

What information do we have:

Ea = 6.30 kJ/mol

A = 6.00×108[tex]M^{-1} .s^{-1}[/tex]

Reaction:

[tex]NO(g) + F2(g)[/tex] → [tex]NOF(g) + F(g)[/tex]

To find,

Rate constant = ?

By employing the equation of Arrhenius,

[tex]k = a[/tex] × [tex]e^{-Ea/R * T}[/tex]

with,

a being the factor of frequency

Ea denoting energy of activation

R denoting the exact constant of gas

T denoting right temperature i.e. 968 K

Temperature Ideal comes from

695°C + 273.15°

= 968 K

by putting the values, we get:

⇒ [tex]2.74[/tex] × [tex]10^8 M^{-1}[/tex][tex].S^{-1}.[/tex] = [tex]2.74[/tex] × [tex]10^8 L.mol^{-1} .s^{-1}[/tex].

Thus, the correct answer is [tex]2.74[/tex] × [tex]10^8 M^{-1}[/tex][tex].S^{-1}.[/tex] = [tex]2.74[/tex] × [tex]10^8 L.mol^{-1} .s^{-1}[/tex].

Learn more about "Frequency" here:

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