🗊 Презентация Rate of reactions. (Chapter 2)

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Rate of reactions. (Chapter 2), слайд №1 Rate of reactions. (Chapter 2), слайд №2 Rate of reactions. (Chapter 2), слайд №3 Rate of reactions. (Chapter 2), слайд №4 Rate of reactions. (Chapter 2), слайд №5 Rate of reactions. (Chapter 2), слайд №6 Rate of reactions. (Chapter 2), слайд №7 Rate of reactions. (Chapter 2), слайд №8 Rate of reactions. (Chapter 2), слайд №9 Rate of reactions. (Chapter 2), слайд №10 Rate of reactions. (Chapter 2), слайд №11 Rate of reactions. (Chapter 2), слайд №12 Rate of reactions. (Chapter 2), слайд №13
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Rate of reactions. (Chapter 2), слайд №1
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Rate of reactions. (Chapter 2), слайд №2
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Rate of reactions. (Chapter 2), слайд №3
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Rate of reactions. (Chapter 2), слайд №4
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The decomposition of dinitrogen pentoxide can be represented by the equation; 2N2O5 → 4NO2 + O2 The concentration of dinitrogen pentoxide decreases...
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The decomposition of dinitrogen pentoxide can be represented by the equation; 2N2O5 → 4NO2 + O2 The concentration of dinitrogen pentoxide decreases from 0,008 M to 0,004 M in 20 seconds. Find the average rate of consumption of dinitrogen pentoxide

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RateN2O5 = (0.008 – 0.004)/20 = 0.0002 = 2.10−4 mol/L. s
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RateN2O5 = (0.008 – 0.004)/20 = 0.0002 = 2.10−4 mol/L. s

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Rate of reactions. (Chapter 2), слайд №7
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Rate of reactions. (Chapter 2), слайд №8
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Rate of reactions. (Chapter 2), слайд №9
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Rate of reactions. (Chapter 2), слайд №10
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Rate of reactions. (Chapter 2), слайд №11
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Temperature The increase in temperature increases rate of reaction. And rate can be calculated by the formula below. Vt2= Vt1. (t2 - t1)/10  is a...
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Temperature The increase in temperature increases rate of reaction. And rate can be calculated by the formula below. Vt2= Vt1. (t2 - t1)/10  is a constant and given in the question. Vt1 = velocity at initial temperature, Vt2 = velocity at final temperature

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Ex: If temperature increases 10 0C, rate increases 3 times. How many 0C must the temperature be increased so that rate can increase 27 times? If t2 -...
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Ex: If temperature increases 10 0C, rate increases 3 times. How many 0C must the temperature be increased so that rate can increase 27 times? If t2 - t1=10 Vt2 = 3Vt1 Then Vt2 = 3 =  10/10 3 =  1 so  = 3 Vt1 If Vt2 = 27.Vt1 then Vt2 = 27 = 3 (t2 - t1)/10 (t2 - t1)/ 10 = 3 Vt1 So t2 - t1 = 30 0C

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