Answer:
radiant - chemical is the best answer
This is a strong base / week acid reaction.
NaOH + CH3COOH
The equilibrium of this reaction is very displaced to the right leading to the formation of the products
Na CH3COO + H2O
Na CH3COOH is a ionic compound which in solutionn will be as Na (+) and CH3COOH(-)
=> CH3COOH + NaOH = CH3 COO(-) + Na(+) + H2O
So, the predominant species in the solution are the ions Na(+) and CH3COO(-).
In general, in an strong base / weak acid titration, the predominant species present at the stoichiometric point will be the cation of the strong base (Na+ in this case) and the conjugate base of the weak acid (the anion of the weak acid, which is CH3COO- in this case).
The answer is predominantly Na(+) and CH3COO(-); predominantly because it is an equlibrium which means that the rectants will also br present.
When an entire species perishes the species becomes extinct.
As given:
Initial moles of P taken = 2 mol
the products are R and Q
at equilibrium the moles of
R = x
total moles = 2 + x/2
Let us check for each reaction
A) P <-> 2Q+R
Here if x moles of P gets decomposed it will give 2x moles of Q and x moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = 2x
moles of R = x
Total moles = (2-x) + 2x + x = 2 +2x
B) 2P <-> 2Q+R
Here x moles of P will give x moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x
moles of R = x/2
Total moles = (2-x) + x + x/2 = 2 + x/2
C) 2P <-> Q+R
Here x moles of P will give x/2 moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x /2
moles of R = x/2
Total moles = (2-x) + x + x = 2
D) 2P <-> Q+2R
Here x moles of P will give x/2 moles of Q and x moles of R
So at equilibrium
moles of P left = 2-x
moles of Q = x/2
moles of R = x
Total moles = (2-x) + x/2 + x = 2 + x/2
