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3 years ago

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The pH of a solution is 8. Some hydrochloric acid is added to the solution. Suggest the pH of the solution after mixing. pH=....

...??

1 answer:

Molodets [167]3 years ago

3 0

Answer:

Probably around 6 because the ph of hydrochloric acid is 3

Explanation:

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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

Artist 52 [7]

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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