by dissolving NaBF4(s) in distilled water. In another experiment, a o.150 MBF4 (
ID: 579016 • Letter: B
Question
by dissolving NaBF4(s) in distilled water. In another experiment, a o.150 MBF4 (a) solution is prepared The BF, (ag) ions in the solution slowly react with H,OU) in the reversible reaction represented below. The concentration of HF is monitored over time, as shown in the graph below. 0.02 = 0.01 0.00 0 200 400 600 800 Time (min) [HF] reaches a constant value of 0.0174 Mwhen the reaction reaches equilibrium. For the forward reacti the rate law is rate= kr[BF41. The value of the rate constant kr was experimentally determined to be 9.00x 10-4 min-1. (b) Calculate the rate of the forward reaction after 600. minutes. Include units with your answer. [BF4 0.150 M-0.0 174 M= 0.133 M rates(9.00 × 10-4 min-1)(0.133 M) = 1.20 × 10-4 Mrnin-l 1 point is earned for the molarit 1 point is earned for the value of rate 1 point is earned for units. The rate law for the reverse reaction is rate= k,[BF30H-][HF]Explanation / Answer
First you have to be clear that the term 'k' is called rate constant, and this is not the same as rate.
Rate of reaction is a quantity which has to be calculated at a particular point of time and may vary for different points of time, but rate constant 'k' is a different thing. It is the proportionality constant in the rate equation and it remains constant throughout.
For the forward reaction, rate equation is:
r = k*[BF4-]
Here, k is given to us and is equal to 9*10-4
We need not calculate 'k' here, we have been asked to calculate 'r' here.
The integrated rate law for a first order reaction is given by:
A = A0*e(-kt).
Here, A0 is the initial conc and A is the conc at time t.
You can not use this equation to calculate the rate 'r', because this equation contains 'k' and not 'r'.
For calculating the rate of reaction 'r', you have to make use of rate law expression only.
You are confusing the two things 'k' and 'r'.
Hope this helps !
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