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3 years ago

11

he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c

onstant of this reaction is at , what will the rate constant be at ?

1 answer:

Leya [2.2K]3 years ago

8 0

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is $61.29M^{-1}s^{-1}$

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{244^oC}$ = equilibrium constant at 244°C = $6.7M^{-1}s^{-1}$

$K_{324^oC}$ = equilibrium constant at 324°C = ?

$E_a$ = Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $244^oC=[273+244]K=517K$

$T_2$ = final temperature = $324^oC=[273+324]K=597K$

Putting values in above equation, we get:

$\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}$

Hence, the rate constant at 324°C is $61.29M^{-1}s^{-1}$

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The solubility of BaCO3(s) in water at a certain temperature is 4.4 10–5 mol/L. Calculate the value of Ksp for BaCO3(s) at this

azamat

Answer : The value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$ .

Solution : Given,

Solubility of $BaCO_3$ in water = $4.4\times 10^{-5}mole/L$

The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.

The equilibrium equation is,

$BaCO_3\rightleftharpoons Ba^{2+}+CO^{2-}_3$

Initially - 0 0

At equilibrium - s s

The Solubility product will be equal to,

$K_{sp}=[Ba^{2+}][CO^{2-}_3]$

$K_{sp}=s\times s=s^2$

$[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L$

Now put all the given values in this expression, we get the value of solubility constant.

$K_{sp}=(4.4\times 10^{-5}mole/L)^2=19.36\times 10^{-10}mole^2/L^2$

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3 years ago

Reversible reaction and examples on them

Tomtit [17]

Answer:

Here are three examples

Explanation:

In a reversible reaction, the conversions of reactants to products and of products to reactants occur at the same time.

Example 1

The reaction of hydrogen and iodine to from hydrogen iodide.

H₂ + I₂ ⇌ 2HI

Example 2

The dissociation of carbonic acid in water to form hydronium and hydrogen carbonate ions

H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃⁻

Example 3

The dissociation of dinitrogen tetroxide to nitrogen dioxide.

N₂O₄ ⇌ 2NO₂

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3 years ago

When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be

Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL + 50.mL = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH $_{neutralization}$

ΔH $_{neutralization}$ = -q_soln / mole of water produced

so we substitute

ΔH $_{neutralization}$ = -( 4076.68 J ) / 0.0500 mol

ΔH $_{neutralization}$ = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

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vichka [17]

Answer:

HF

H₂S

H₂CO₃

NH₄⁺

Explanation:

<em>Which acid in each of the following pairs has the stronger conjugate base?</em>

According to Bronsted-Lowry acid-base theory, <em>the weaker an acid, the stronger its conjugate acid</em>. Especially for weak acids, pKa gives information about the strength of such acid. <em>The higher the pKa, the weaker the acid.</em>

<em />

Of the acids HCl or HF, the one with the stronger conjugate base is HF because it is a weak acid.

Of the acids H₂S or HNO₂, the one with the stronger conjugate base is H₂S because it is a weaker acid. pKa (H₂S) = 7.04 > pKa (HNO₂) = 3.39

Of the acids H₂CO₃ or HClO₄, the one with the stronger conjugate base is H₂CO₃ because it is a weak acid.

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3 years ago

Which list ranks the gasses in the correct order from the

JulijaS [17]

Answer:

D. chlorine, oxygen, nitrogen, hydrogen.

Explanation:

Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.

<em>ν ∝ 1/√M</em>

where ν is the rate of effusion and M is the atomic or molecular mass of the gas particles.

The molecular mass for the listed gases are:

O₂: 32.0 g/mol,

Cl₂: 70.906 g/mol,

N₂: 28.0 g/mol,

H₂: 2.0 g/mol.

Hence, the smallest molecular mass of the gas, the fastest rate of effusion.

So, the order from the slowest to the fastest rate of effusion is:

<em>Chlorine, oxygen, nitrogen, hydrogen.</em>

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3 years ago

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