Scanned by CamScanner Scanned by CamScanner Scanned by CamScanner Scanned by CamScanner Scanned by CamScanner Scanned by CamScanner Part II Data collection 6. Mixture 1 2 3 4 Time(s) 190 99 144 58 5 163 Rate(M/s) 1.05 × 10−6 2.02 × 10−6 1.39 × 10−6 3.45 × 10−6 1.23 × 10−6 Concentration After Mixing [I-] [BrO3-] 0.0020 0.0080 0.0040 0.0080 0.0020 0.0160 0.0020 0.0080 [H+] 0.02 0.02 0.02 0.04 0.0018 0.03 0.0048 7. 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆2 𝑂32− = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 × 𝑖𝑛𝑡𝑖𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝐹𝑖𝑛𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒 Considering trial 1 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑆2 𝑂32− = 0.001 = 0.0002 𝑀 5 At the all trials volume of the 𝑆2 𝑂32−did not change. Therefore, 𝑆2 𝑂32− for all trials = 0.0002M 8. Considering trial 1 ∆[𝑆2 𝑂32− ] 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = ∆𝑡 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 0.0002𝑀 = 2.02 × 10−6 𝑀/𝑠 99𝑠 9. Considering trial 1 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 H+ = 0.1 × 10𝑚𝑙 = 0.02𝑀 50𝑚𝑙 Data Analysis, 1. 𝑟 = −𝑘[𝐼 − ]𝑎 [𝐵𝑟𝑂3− ]𝑏 [𝐻 + ]𝑐 2. I would choose 1st and 2nd experiment to find the exponent of the 𝐼 − . Because In those experiments the concentration of 𝐵𝑟𝑂3− and 𝐻 + remain unchanged while the concentration of 𝐼 − is changing. 3. 𝑟 = 𝑘[𝐼 − ]𝑎 [𝐵𝑟𝑂3− ]𝑏 [𝐻 + ]𝑐 1st experiment 10−6 𝑀 1.05 × = 𝑘[0.0020]𝑎 [0.0080]𝑏 [0.02]𝑐 − − − − − −(1) 𝑠 2nd experiment 2.02 × 10−6 𝑀 = 𝑘[0.0040]𝑎 [0.0080]𝑏 [0.02]𝑐 − − − − − −(2) 𝑠 (1)/(2) 10−6 𝑀 𝑘[0.0020]𝑎 [0.0080]𝑏 [0.02]𝑐 𝑠 = 10−6 𝑀 𝑘[0.0040]𝑎 [0.0080]𝑏 [0.02]𝑐 2.02 × 𝑠 1.05 × 1 1 𝑎 =( ) 1.9238 2 𝑎=1 4. I would choose 1st and 3rd experiment to find the exponent of the 𝐵𝑟𝑂3− . Because In those experiments the concentration of 𝐼 − and 𝐻 + remain unchanged while the concentration of 𝐵𝑟𝑂3− is changing. 5. 1st experiment 10−6 𝑀 1.05 × = 𝑘[0.0020]𝑎 [0.0080]𝑏 [0.02]𝑐 − − − − − −(1) 𝑠 3rd experiment 1.39 × 10−6 𝑀 = 𝑘[0.0040]𝑎 [0.0160]𝑏 [0.02]𝑐 − − − − − −(3) 𝑠 (1)/(3) 1.05 0.008 𝑏 =( ) 1.39 0.016 𝑏 = 0.4 𝑏≈1 6. I would choose 1st and 4th experiment to find the exponent of the 𝐻 + . Because In those experiments the concentration of 𝐼 − and 𝐵𝑟𝑂3− 𝐻+ remain unchanged while the concentration of 𝐻 + is changing. 7. 1st experiment 1.05 × 10−6 𝑀 = 𝑘[0.0020]𝑎 [0.0080]𝑏 [0.02]𝑐 − − − − − −(1) 𝑠 3.45 × 10−6 𝑀 = 𝑘[0.0020]𝑎 [0.0080]𝑏 [0.04]𝑐 − − − − − −(4) 𝑠 4th experiment (1)/(4) 1.05 1 𝑐 =( ) 3.45 2 𝑐≈2 Data analysis 1. 𝑟 = −𝑘[𝐼 − ]1 [𝐵𝑟𝑂3− ]1 [𝐻 + ]2 2. 𝒌𝟏 = 𝟏𝟔𝟒. 𝟎𝟔𝟐𝟓/𝑴𝟑 𝒔 𝒌𝟐 = 𝟏𝟓𝟕. 𝟖𝟏𝟐𝟓/𝑴𝟑 𝒔 𝒌𝟑 = 𝟏𝟎𝟖. 𝟓𝟗𝟑𝟕/𝑴𝟑 𝒔 𝒌𝟒 = 𝟏𝟑𝟒. 𝟕𝟔𝟓𝟔/𝑴𝟑 𝒔 Average = 𝟏𝟒𝟏. 𝟑𝟎𝟖𝟔/𝑴𝟑 𝒔 Data Prediction 1. 𝑟 = −𝑘[𝐼 − ]1 [𝐵𝑟𝑂3− ]1 [𝐻 + ]2 𝑟 = 141.3086(0.0018)(0.0048)(0.03)2 𝑟 = 1.01 × 10−6 𝑀/𝑠 ∆𝑡 = 0.0002𝑀 1.01 × 10−6 𝑀/𝑠 ∆𝑡 = 198 𝑠 2. Actual value = 163 s 𝐸𝑟𝑟𝑜𝑟 = 198 − 163 × 100% = 17.68% 198 This may happen because fail to recognize the exact color changing point. Questions 1. When temperature increase the rate of any reaction will increase. Therefore, the observed reaction rate may not the actual reaction rate. The observations give higher reaction rate than expected. This may cause to the calculations of the exponents values of the reactant in the rate law equation. 2. The rate law is related with the slowest reaction. But the reaction mechanism should include all reactant of this experiment. Because all of the reactant are represented in the rate reaction.
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