For a certain reaction, kc = 2.76×103 and kf= 6.54×10−4 m−2⋅s−1 . calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction. express your answer with the appropriate units. include explicit multiplication within units, for example to enter m−2⋅s−1 include ⋅ (multiplication dot) between each measurement.

2.37x10⁻⁷ M⁻².s⁻¹

Explanation:

For a generic reversible reaction:

A + B ⇄ C + D

Kf is the constant of the formation of the products (C and D), Kr is the constant of the formation of the reactants (A and B), and Kc is the general equilibrium constant, which is:

Kc = Kf/Kr

2.76x10³ = 6.54x10⁻⁴/Kr

Kr = 6.54x10⁻⁴/2.76x10³

Kr =  2.37x10⁻⁷ M⁻².s⁻¹

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.