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2 years ago

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1. The hydrogenation of ethene gas at 298 K shows a decrease in disorder (AS-120.7 J/mol-K) during an exothermic reaction ( Delt

a H^ =-136.9 k/mol . Determine whether the reaction is spontaneous or nonspontaneous by calculating Delta * 0 degrees

1 answer:

Anna71 [15]2 years ago

8 0

K shows a decrease in disorder (AS = -120.7 J/(mol-K)) during an exothermic reaction (AH' = -136.9 kJ/mol). Determine whether the reaction is spontaneous or ...

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Pure substance definition in chemistry?

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What is the pH of a 0.50 M C6H5NH3Br solution? KbC6H5NH2 = 3.9x10-10 (R = 2.45)

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Explanation:

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An empty beaker is weighed and found to weigh 23.1 g. Some potassium chloride is then added to the beaker and weighed again. The

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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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3 years ago