2 C2F4 → C4F8 is 0.0410 M−1 s −1 . We start with 0.105 mol C2F4 in a 4.00-liter container, with no C4F8 initially present. What will be the concentration of C2F4 after 3.00 hours ? Answer in units of M.

After three hours, concentration of C₂F₄ is 0.00208M

Explanation:

As the units of the rate constant of the reaction are M⁻¹s⁻¹ we can know this reaction is of second order. Its integrated law is:

Where [A] and [A]₀ represents initial and final concentrations of the reactant (C₂F₄), K is rate constant (0.0410M⁻¹s⁻¹) and t is time.

3.00 hours are in seconds:

3 hours ₓ (3600 seconds / 1 hour) = 10800 seconds

Computing in the equation:

[A] = 0.00208M

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

your answer would be “dry”