What is the concentration in molarity of S2O32- (aq) in a solution prepared by mixing 150 mLmL of 0.149 MM Na2S2O3 (aq) with eno
1 answer:
Answer:
0.0890 M
Explanation:
Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:
C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL
C2 = 0.149 x 150/250
= 0.089 M
To determine the concentration of S2O32- (aq), consider the equation:
The concentration of Na2S2O3 and S2O32- (aq) is 1:1
Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.
To 3 significant figures = 0.0890 M
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Answer : The order of reaction with respect to A is, first order reaction.
The order of reaction with respect to B is, zero order reaction.
The overall order of reaction is, first order reaction.
Explanation :
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
Rate law expression for the reaction:
where,
a = order with respect to A
b = order with respect to B
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Expression for rate law for third observation:
....(3)
Dividing 1 from 2, we get:
Dividing 1 from 3, we get:
Thus, the rate law becomes:
Thus,
The order of reaction with respect to A is, first order reaction.
The order of reaction with respect to B is, zero order reaction.
The overall order of reaction is, first order reaction.