Consider the reaction N2(g) + 3H2(g) → 2NH3(g) Suppose that at a particular moment during the reaction, molecular hydrogen is reacting at the rate of −0.091 M/s.
(a) At what rate is ammonia being formed? M/s
(b) At what rate is molecular nitrogen consumed? M/s

The solution to the question is as follows

(a) The rate of ammonia formation = 0.061 M/s

(b) the rate of N₂ consumption = 0.0303 M/s

Explanation:

(a) To solve the question we note that the reaction consists of one mole of N₂ combining with three moles of H₂ to form 2 moles of NH₃

N₂(g) + 3H₂(g) → 2NH₃(g)

The rate of reaction of molecular hydrogen = 0.091 M/s, hence we have

3 moles of H₂ reacts to form 2 moles of NH₃, therefore

0.091 M of H₂ will react to form 2/3 × 0.091 M or 0.061 M of NH₃

Hence the rate of ammonia formation is 0.061 M/s

(b) From the reaction equation we have 3 moles of H₂ and one mole of N₂ being consumed at the same time hence

0.091 M of H₂ is consumed simultaneously with 1/3 × 0.091 M or 0.0303 M of N₂

Therefore the rate of consumption of N₂ = 0.0303 M/s

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.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"