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3 years ago

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The specific reaction rate of a reaction at 492k is 2.46 second inverse and at 528k 47.5 second inverse.find activation energy a

nd ko.

1 answer:

crimeas [40]3 years ago

3 0

Answer:

Ea = 177x10³ J/mol

ko = $1.52x10^{19}$ J/mol

Explanation:

The specific reaction rate can be calculated by Arrhenius equation:

$k = koxe^{-Ea/RT}$

Where k0 is a constant, Ea is the activation energy, R is the gas constant, and T the temperature in Kelvin.

k depends on the temperature, so, we can divide the k of two different temperatures:

$\frac{k1}{k2} = \frac{koxe^{-Ea/RT1}}{koxe^{-Ea/RT2}}$

$\frac{k1}{k2} = e^{-Ea/RT1 + Ea/RT2}$

Applying natural logathim in both sides of the equations:

ln(k1/k2) = Ea/RT2 - Ea/RT1

ln(k1/k2) = (Ea/R)x(1/T2 - 1/T1)

R = 8.314 J/mol.K

ln(2.46/47.5) = (Ea/8.314)x(1/528 - 1/492)

ln(0.052) = (Ea/8.314)x(-1.38x $10^{-4}$

-1.67x $10^{-5}$ xEa = -2.95

Ea = 177x10³ J/mol

To find ko, we just need to substitute Ea in one of the specific reaction rate equation:

$k1 = koxe^{-Ea/RT1}$

$2.46 = koxe^{-177x10^3/8.314x492}$

$1.61x10^{-19}ko = 2.46$

ko = $1.52x10^{19}$ J/mol

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Answer:

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Putting values in equation 1, we get:

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$NH_3+HCl\rightarrow NH_4^++Cl^-$

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1 mole of HCl reacts with 1 mole of ammonia

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Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M$

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1 mole of ammonia produces 1 mole of ammonium ion

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Moles of ammonium ion = 0.0035 moles

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Putting values in equation 1, we get:

$\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M$

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$K_w=K_b\times K_a$

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$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.8\times 10^{-5}$

$10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

The chemical equation for the dissociation of ammonium ion follows:

$NH_4^+\rightarrow NH_3+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

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$[NH_3]=[H^+]=x$

$[NH_4^+]=0.072M$

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