Using the Arrhenius Equation, what is the rate constant (k) for the reaction of NO with F2 at a temperature of 298 K, given that the activation energy (Ea) is 6.30 kJ/mol and the frequency factor (A) is 6.00 × 10^8 M⁻¹·s⁻¹?

• A. k = 5.67 × 10^9 M⁻¹·s⁻¹
• B. k = 4.56 × 10^8 M⁻¹·s⁻¹
• C. k = 1.23 × 10^7 M⁻¹·s⁻¹
• D. k = 2.45 × 10^7 M⁻¹·s⁻¹

Answer: (D)

Verified step by step guidance

1. Start by recalling the Arrhenius Equation, which is used to calculate the rate constant (k) of a reaction: \( k = A \cdot e^{-\frac{E_a}{RT}} \), where \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
2. Convert the activation energy \( E_a \) from kJ/mol to J/mol to match the units of the gas constant \( R \). Since 1 kJ = 1000 J, multiply 6.30 kJ/mol by 1000 to get 6300 J/mol.
3. Use the universal gas constant \( R = 8.314 \) J/(mol·K) and the given temperature \( T = 298 \) K in the Arrhenius equation.
4. Substitute the values into the Arrhenius equation: \( k = 6.00 \times 10^8 \cdot e^{-\frac{6300}{8.314 \times 298}} \).
5. Calculate the exponent \( -\frac{6300}{8.314 \times 298} \), then find the value of \( e \) raised to this power, and finally multiply by the frequency factor \( A = 6.00 \times 10^8 \) to find the rate constant \( k \).