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alexira [117]
3 years ago
13

In a first–order reaction (A → B and rate = k [A]), tripling the concentration of A will have what effect on the reaction rate?

Chemistry
1 answer:
Natali [406]3 years ago
5 0

Answer:

Tripling the concentration of A will triple the reaction rate.

Explanation:

  • For a first–order reaction hat has a rate law:

<em>Rate = k[A].</em>

  • It is clear that the reaction rate is directly proportional to the concentration of A.

<em>Rate ∝ [A].</em>

<u><em>So, Tripling the concentration of A will triple the reaction rate.</em></u>

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Propane gas, C3H8, is sometimes used as a fuel. In order to measure its energy output as a fuel a 1.860 g sample was combined wi
lisov135 [29]

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

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Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ

Moles of propane =\frac{1.860 g}{44 g/mol}=0.0422 mol

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol

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