For the reaction, A(g) + B(g) => AB(g), the rate is 0.625 mol/L. s when the initial concentrations of both A and B are 2.00 mol/L. If the reaction is first order in A and first order in B, what is the rate when the initial concentration of A = 4.41 mol/L and that of B = 2.72 mol/L. Give answer to 2 decimal places.

The rate of concentration is 1.87 mol L⁻¹s⁻¹

Explanation:

First order reaction: First order reaction is only depends on the value of the concentration of only one reactant.

A(g)+B(g)⇒AB(g)

rate of reaction = k [A][B]   [ since A and B are first order reaction]

k= rate constant

[A] = concentration of A

[B] = concentration of B

Given that rate = 0.625 mol/L.s

[A] = 2 mol/L   and   [B] = 2 mol/L

Therefore,

rate of reaction = k [A][B]

⇒0.625= k ×2×2

The value of k changes only when the temperature changes.

Now [A]= 4.41 mol/L    and   [B]=  2.72 mol/L

Rate of reaction = K[A][B]

 ≈1.87 mol L⁻¹s⁻¹

The rate of concentration is 1.87 mol L⁻¹s⁻¹

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.