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likoan [24]
3 years ago
12

Which of these statements is not true about chemical reaction rates? a Temperature can change a reaction rate. b The amount of r

eactants can increase the reaction rate. c Temperature can decrease the reaction rate. d Temperature does not affect the reaction rate.
Chemistry
1 answer:
Umnica [9.8K]3 years ago
4 0

Answer:

The statement that is not true is: 'Temperature does not affect the reaction rate'.  

Explanation:

a) Temperature can change a reaction rate. <u> This is true</u>

Increasing the temperature increases the reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

For example, the time taken to melt a metal will be much higher at a lower temperature but it will decrease as soon as we increase the temperature

b) The amount of reactants can increase the reaction rate.<u> This is true</u>

A higher concentration of reactants leads to more effective collisions per unit time, which leads to an increased reaction rate.

c) Temperature can decrease the reaction rate. <u>This is true </u>

Decreasing the temperature decreases the reaction rates because of the  decrease in the number of high energy collisions. It will result in a slower reaction.

d) Temperature does not affect the reaction rate.  <u>This is not true. </u>

The reaction rate is temperature dependent. The reaction rate increases with higher temperature and decreases with lower temperature.

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A solution has [c6h5cooh] = 0.100 m and [ca(c6h5coo)2] = 0.200 m. ka = 6.3 × 10−5 for c6h5cooh. the solution volume is 5.00 l. w
Eddi Din [679]

6.8 is the pH of the solution after 10 ml of 5M NaOH is added.

Explanation:

Data given:

Molarity of C6H5CCOH = 0.100 M

molarity of ca(c6h5coo)2  = 0.2 M

Ka = 6.3 x 10^-5

first pH is calculated of the buffer solution

pH = pKa+ log 10 \frac{[A-]}{[HA]}

pKa = -log10[Ka]

pka = -log[6.3 x10^-5]

pKa = 4.200

putting the values to know pH of the buffer

pH = 4.200 + log 10 \frac{0.2}{0.1}

pH = 4.200 + 0.3

    pH  = 4.5 (when NaOH was not added, this is pH of buffer solution)

now the molarity of the solution is calculated after NaOH i.e Mbuffer is added

MbufferVbuffer = Mbase Vbase

putting the values in above equation:

Mbuffer = \frac{MbaseVbase}{Vbuffer}

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