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Cerrena [4.2K]
3 years ago
13

The first order reaction of methyl isonitrile to form acetonitrile has a rate constant of k = 1.7 × 10−15 at 298 K. What will ha

ppen to the rate constant if the temperature is decreased to 100 K?

Chemistry
2 answers:
tino4ka555 [31]3 years ago
5 0

Answer : If the temperature is decreased to 100 K then the rate constant will be, decrease.

Explanation :

According to the Arrhenius equation,  the rate constant of the reaction is directly proportional to the temperature.

K=a\times e^{\frac{-Ea}{RT}}

Taking ln on both the sides, we get

\ln K=\frac{-Ea}{RT}+\ln a

where,

K = rate constant

Ea = activation energy

T = temperature

R = gas constant

a = Arrhenius constant

As we know that the increase in the rate of reaction with increase in the temperature is mainly due to increase in the number of effective collision. That means as the temperature increases, the number of effective collision increases and the value of rate constant also increases or vice-versa.

As per question, the temperature of the reaction decreases that means the number of effective collision decreases and the value of rate constant also decreases.

Hence, if the temperature is decreases from 298 K to 100 K then the rate constant will be, decrease.

ludmilkaskok [199]3 years ago
3 0
Refer to to this chart 

Credits to <span>https://image.slidesharecdn.com/nybf09-unit2slides26-57-100125181507-phpapp01/95/nyb-f09-unit-2-slid...

As temperature decreases, rate decreases. 

As temp increases, k also increases. </span>

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3 years ago
If 54 g of Al reacted with 160 g of O2, find out the weight of product​
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A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc
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Answer : The volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of O_2 gas = (740-22.4) torr = 717.6 torr

P_2 = final pressure of O_2 gas at STP= 760 torr

V_1 = initial volume of O_2 gas = 280 mL

V_2 = final volume of O_2 gas at STP = ?

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Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

V_2=242.2mL=0.2422L

Therefore, the volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

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