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The rate of a certain reaction doubles for every 10°C rise in temperature.

(a) How much faster does the reaction proceed at 45°C than at 25°C?

(b) How much faster does the reaction proceed at 95°C than at 25°C?

Short Answer

Expert verified

The rate increases by four times.

The rate increases by 128 times.

Step by step solution

01

Calculating the difference in temperature

For every10°C rise in temperature the rate is 2 times the original.

Difference in temperature =\({\bf{(45 - 25)^\circ C = 20^\circ C}}\)

02

Step 2: Difference in rate

Thus, for 20°C rise in temperature the rate increases by \({{\bf{(2)}}^{\bf{2}}}{\bf{ = 4}}\)

03

Calculating the difference in temperature

Difference in temperature =\({\bf{(95 - 25)^\circ C = 70^\circ C}}\)

04

Step 4: Difference in rate 

Thus for \(70^\circ C\)rise in temperature the rate increases by \({{\bf{(2)}}^{\bf{7}}}{\bf{ = 128}}\) .

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Most popular questions from this chapter

Does the following data fit a second-order rate law?

Trial

Time(s)

(A) (M)

1

5

0.952

2

10

0.625

3

15

0.465

4

20

0.370

5

25

0.308

6

35

0.230

How much and in what direction will each of the following effect the rate of the reaction:

CO(g) + \({\bf{NO}}{}_{\bf{2}}\) (g)⟶ \({\bf{CO}}{}_{\bf{2}}\) (g) + NO(g) if the rate law for the reaction is rate =\({\bf{k(NO}}{}_{\bf{2}}{{\bf{)}}^{\bf{2}}}{\bf{a}}\)?

  1. Decreasing the pressure of \({\bf{NO}}{}_{\bf{2}}\) from 0.50 atm to 0.250 atm.
  2. Increasing the concentration of CO from 0.01 M to 0.03 M.

:How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate

The annual production of \({\bf{HN}}{{\bf{O}}_{\bf{3}}}\) in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.

\(\begin{align}\left( a \right){\bf{ }}4N{H_3}{\bf{ }}\left( g \right){\bf{ }} + {\bf{ }}5{O_2}{\bf{ }}(g) \to 4NO\left( g \right){\bf{ }} + {\bf{ }}6{H_2}O\left( g \right)\\\left( b \right){\bf{ }}2NO\left( g \right){\bf{ }} + {\bf{ }}{O_{2{\bf{ }}}}(g) \to 2N{O_{2{\bf{ }}}}\left( g \right)\\\left( c \right){\bf{ }}3N{O_2}{\bf{ }}\left( g \right){\bf{ }} + {\bf{ }}{H_2}O(l) \to 2HN{O_3}(aq) + NO(g)\end{align}\)

The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in \({{\bf{O}}_{\bf{2}}}\), what is the rate of formation of \({\bf{N}}{{\bf{O}}_{\bf{2}}}\) when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is \({\bf{5}}{\bf{.8 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ L}}{{\bf{ }}^{\bf{2}}}{\bf{ mo}}{{\bf{l}}^{{\bf{ - 2}}}}{\bf{ s}}{{\bf{ }}^{{\bf{ - 1}}}}\).

Under certain conditions, the decomposition of ammonia on a metal surface gives the following data:

Determine the rate law, the rate constant, and the overall order for this reaction.

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