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kobusy [5.1K]
3 years ago
5

In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in

the solution slowly react with H2O(l) in the reversible reaction represented below.
BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

[HF] reaches a constant value of 0.0174 M when the reaction reaches equilibrium. For the forward reaction, the rate law is rate = kf [BF4-]. The value of the rate constant kf was experimentally determined to be 9.00x10-4 min-1.

Required:
a. Calculate the rate of the forward reaction after 600 minutes.
b. A student claims that the initial rate of the reverse reaction is equal to zero. Do you agree or disagree with this claim?
Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0

Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

[BF4-] = 0.1326 M

From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

Forward rate = 9.00 × 10^(-4) /min × (0.1326M)

Forward rate = 1.1934 × 10^(-4) M/min

(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

Thus, I disagree with the claim.

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