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AleksandrR [38]
3 years ago
10

Watered down koolaid is an example of

Chemistry
1 answer:
N76 [4]3 years ago
5 0
The answer is A . Saturated solution
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The initial reaction rate for the elementary reaction 2A + B → 4C was measured as a function of temperature when the concentrati
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Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The activation energy is 124.776\frac{kJ}{mole}

b) The frequency factor is 1.77 ×10^{18}

c) The rate constant is 0.00033 (\frac{dm^{3} }{mole} )^{2}\frac{1}{s}

Explanation:

From the question the elementary reaction for A and B is given as

      2A + B → 4C

The rate equation the elementary reaction is

      -r_{A} = k[A]^{2}[B]

            =  k[2]^{2}[1.5]

            = 6k

     k = \frac{-r_{A} }{6}

      When temperature changes, the rate constant change an this causes the rate of reaction to change as shown on the second uploaded image.

The relationship between temperature and rate constant can be deduced from these equation

                    k = Aexp(-\frac{E_{a} }{RT} )

             taking ln of both sides we have

                   lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}

        Considering the graph for the rate constant ln k and (\frac{1}{T} ) the slope from the equation is -(\frac{E_{a} }{R}) and the intercept is ln A

From the given table we can generate another table using the equation above as shown on the third uploaded image

The graph of ln k  vs (\frac{1}{T} )  is shown on the fourth uploaded image

  From the graph we can see that the slope is -(\frac{E_{a} }{R} ) = - 15008

Now we can obtain the activation energy E_{a} by making it the subject in the equation also generally R which is the gas constant is 8.145 \frac{J}{kmole}

                E_{a}  = 15008 ×  8,3145\frac{J}{molK}  

                     = 124\frac{KJ}{mole}

    Hence the activation energy is = 124\frac{KJ}{mole}

b) From the graph its intercept is ln A = 42.019

                                                          A = exp(42.019)

                                                             =1.77 × 10^{18}

Hence the frquency factor A is  =1.77 × 10^{18}

c) From the equation of rate constant

                                          lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}

We have

                ln k = 42.019 - 15008 * (\frac{1}{300} )

                      k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}

Hence the rate constant is k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}    

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