Calculate the intial reaction rate for all five runs , refer to this equation k=

Question

Calculate the intial reaction rate for all five runs , refer to this equation k=-delta[H2O2]/delta time =-1/2delta [S2O3]/delta time

Conc of H2O2:0.2049+-0.0002 M

Conc of Na2S2O3: 0.0225+-0.0002 M

run1 initial reading final reading volume added time H202 2.11 ml 12.12 ml 10.01 ml 4:29 mins Na2S2O3 6.30 ml 16.35 ml 10.05 ml temperature for run 1: 21 degrees H2O 0.90 ml 21.10 ml 20.05 ml Run2 initial reading final vol. added time H2O2 42.21 ml 62.22 ml 20.00 ml 2:15 min Na2S2O3 17.28 ml 27.50 ml 10.22 ml temperature for run 2: 21.30 degrees H2O 31.32 ml 41.40 ml 10.08 ml

Explanation / Answer

(1)

Ini Conc of H2O2 0.2049+-0.0002 M

Ini Conc of Na2S2O3: 0.0225+-0.0002 M

So for H2O2

M1V1=M2V2

V1=2.11 ml

V2=12.12 ml

M2=M1V1/V2=(0.2049+-0.0002)*2.11/12.12=0.03567+-0.000035

t=4:29 min=269 sec

Initial rates=-delta[H2O2]/delta t=-(0.03567+-0.000035-0.2049+-0.0002)/269=0.00063+-0.00000006

Run 2:

M2=(0.2049+-0.0002)*42.21/62.22=0.139+-0.00014

t=135 sec

Initial rates=-delta[H2O2]/delta t=-(0.139+-0.00014-0.2049+-0.0002)/135=0.00048+-0.00000004

Run 3:

M2=(0.2049+-0.0002)*62.22/92.30=0.138+-0.00013

t=95 sec

Initial rates=-delta[H2O2]/delta t=-(0.138+-0.00013-0.2049+-0.0002)/95=0.0007+-0.00000007

Run 4:

M2=(0.2049+-0.0002)*34.99/65.01=0.110+-0.0001

t=180 sec

Initial rates=-delta[H2O2]/delta t=-(0.110+-0.0001-0.2049+-0.0002)/180=0.00053+-0.00000005

Run 5:

M2=(0.2049+-0.0002)*65.01/95.03=0.140+-0.00014

t=9 sec

Initial rates=-delta[H2O2]/delta t=-(0.140+-0.00014-0.2049+-0.0002)/9=0.0072+-0.00000067

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