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Cerrena [4.2K]
4 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]4 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]4 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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Read the given equation. 2Na + 2H2O ? 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with
emmasim [6.3K]

Answer:

The number of moles of Na metal that used initially = 0.70 mol.

The quantity of Na metal used initially to produce 7.80 of H₂ gas = 16.02 g.

Explanation:

  • It is a stichiometry problem.

<em>2Na + 2H₂O → 2NaOH + H₂,</em>

  • The balanced equation shows that <em>2.0 moles of Na metal </em>react with 2.0 moles of water to produce 2.0 moles of NaOH and <em>1.0 mole of H₂</em>,
  • Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: <em>PV = nRT</em>,

where, P is the pressure of the gas in atm<em> (P at STP = 1.0 atm)</em>,

V is the volume of the gas in L <em>(V = 7.80 L)</em>,

n is the number of moles in mole,

R is the general gas constant<em> (R = 0.082 L.atm/mol)</em>,

T is the temperature of the gas in K <em>(T at STP = 0.0 °C + 273 = 273.0 K)</em>.

∴ The number of moles of H₂ gas (n) = PV / RT = [(1.0 atm)(7.80 L)] / [(0.082 L.atm/mol.K)(273.0 K)] = 0.35 mol.

<em>Using cross multiplication:</em>

2.0 moles of Na will produce → 1.0 mole of H₂, from the stichiometrey.

??? moles of Na will produce → 0.35 mole of H₂.

∴ The number of moles of Na metal that used initially = (2.0 mol)(0.35 mol) / (1.0 mol) = 0.70 mol.

Now, we can get the quantity of Na metal using the relation:

∴ mass = n x molar mass = (0.70 mol)(22.989 g/mol) = 16.02 g.

6 0
3 years ago
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