The Bromination of acetone lab - calculating the order of reaction Data for vary

Question

The Bromination of acetone lab - calculating the order of reaction
Data for varying bromine:
Run 1: 0.05 M Br2. 0.25 M HCl. 1 M Acetone Run 2: 0.025 M Br2. 0.25 M HCl. 1M Acetone Run 3: 0.0125 M Br2. 0.25 M HCl. 1 M Acteone
On the graph of [Br2] vs time, the absorbance was recorded for the peak:
1) 0.05 M Br2 At 0.70 seconds. A=1.8021 A 2) 0.025 M Br2. At 0.94 seconds. A=1.2744 A 3) 0.0125 M Br2. At 2.39 seconds. A=0.30275 A
I know I should be using:
Rate 2 = k [acetone]^x [bromine]^y [HCl]^z all divided by
Rate 1 = k [acetone]^x [bromine]^y [HCl]^z
And the k should cancel since it's a constant, concentration of acetone and HCl were kept constant and should cancel
Question: how to find the rate for each and then proceed with finding the exponent "y". Or a better way to do this, because I know I can repeat the process to find the "x" and "z" when I varied the other concentrations.
I'm not sure how graphing the concentration versus time to get the slope will help ? I know it should give me -k for a zero order. I read online that the rate = -slope , can this be an alternative to finding the rate for each? The Bromination of acetone lab - calculating the order of reaction
Data for varying bromine:
Run 1: 0.05 M Br2. 0.25 M HCl. 1 M Acetone Run 2: 0.025 M Br2. 0.25 M HCl. 1M Acetone Run 3: 0.0125 M Br2. 0.25 M HCl. 1 M Acteone
On the graph of [Br2] vs time, the absorbance was recorded for the peak:
1) 0.05 M Br2 At 0.70 seconds. A=1.8021 A 2) 0.025 M Br2. At 0.94 seconds. A=1.2744 A 3) 0.0125 M Br2. At 2.39 seconds. A=0.30275 A
I know I should be using:
Rate 2 = k [acetone]^x [bromine]^y [HCl]^z all divided by
Rate 1 = k [acetone]^x [bromine]^y [HCl]^z
And the k should cancel since it's a constant, concentration of acetone and HCl were kept constant and should cancel
Question: how to find the rate for each and then proceed with finding the exponent "y". Or a better way to do this, because I know I can repeat the process to find the "x" and "z" when I varied the other concentrations.
I'm not sure how graphing the concentration versus time to get the slope will help ? I know it should give me -k for a zero order. I read online that the rate = -slope , can this be an alternative to finding the rate for each?
Data for varying bromine:
Run 1: 0.05 M Br2. 0.25 M HCl. 1 M Acetone Run 2: 0.025 M Br2. 0.25 M HCl. 1M Acetone Run 3: 0.0125 M Br2. 0.25 M HCl. 1 M Acteone
On the graph of [Br2] vs time, the absorbance was recorded for the peak:
1) 0.05 M Br2 At 0.70 seconds. A=1.8021 A 2) 0.025 M Br2. At 0.94 seconds. A=1.2744 A 3) 0.0125 M Br2. At 2.39 seconds. A=0.30275 A
I know I should be using:
Rate 2 = k [acetone]^x [bromine]^y [HCl]^z all divided by
Rate 1 = k [acetone]^x [bromine]^y [HCl]^z
And the k should cancel since it's a constant, concentration of acetone and HCl were kept constant and should cancel
Question: how to find the rate for each and then proceed with finding the exponent "y". Or a better way to do this, because I know I can repeat the process to find the "x" and "z" when I varied the other concentrations.
I'm not sure how graphing the concentration versus time to get the slope will help ? I know it should give me -k for a zero order. I read online that the rate = -slope , can this be an alternative to finding the rate for each?

Explanation / Answer

Yes. Determining slope from the graph will evaluate the rate. And from rate the x, y and z can be determined.

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