The rate constant doubles when the temperature is increased from 45.0 C to 73.0 C. What is the activation energy for this reaction? (R=8.314 J/Kmol)?

Answers

Answer 1
Answer:

Answer:

Ea =22542.6

Explanation:

The rate constant k is affected by the temperature and this dependence may be represented by the Arrhenius equation:

k=Ae^{-(E_a)/(RT) }

where the pre-exponential factor A is assumed to be independent of temperature, R is the gas constant, and T the temperature in K. Taking the natural logarithm of this equation gives:

ln  k = ln A - Ea/(RT)

or

ln  k = -Ea/(RT) + constant

or

ln  k = -(Ea/R)(1/T) + constant

These equations indicate that the plot of ln k vs. 1/T is a straight line, with a slope of -Ea/R. These equations provide the basis for the experimental determination of Ea.

now applying the above equation in the problem

we can write that

ln(k_2)/(k_1) = (E_a)/(R)[(1)/(T_1)-(1)/(T_2)  ]

solve for Ea:

Ea = R[Ln(k2/k1)] / [(1/T1) - (1/T2)]

but k_2 = 2 k_1, hence:

Ea = (8.314 J/moleK)[ln(2)] / [(1/273+45) - (1/273+73)]

Ea =22542.6

Answer 2
Answer:

Answer:

The activation energy for this reaction is 22.6 kJ/ mol

Explanation:

Step 1: Data given

Rate constant doubles when Temperature goes from 45.0 °C to 73.0 °C

R = 8.314 J/K*mol

Step 2: Calculate the activation energy

Log (k2/k1) = Ea / 2.303R *((1/T1) - (1/T2))

⇒ with k1 = initial rate constant

⇒ with k2 = rate constant after doubled = 2k1

⇒ T1 = initial temperature = 45.0 °C = 318 Kelvin

⇒ T2 = Final temperature = 73.0 °C = 346 Kelvin

log (2) = Ea / (2.303*8.314) *((1/318) - (1/346))

log(2) = Ea / (2.303*8.314) * 0.00025448

Ea = 22649 J/mol = 22.6 kJ/mol

The activation energy for this reaction is 22.6 kJ/ mol


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Answers

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Answers

Answer:

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H2O

Answers

Answer:

C6H6 has the strongest intermolecular force

i think

Answer:

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Explanation:

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Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate. The complete ionic equation contains which of the following species (when balanced in standard form)? A. NO (aq)
B. 2Ba (aq)
C. 2Ag (aq)
D. CI(aq)

Answers

Answer:

Option C

Explanation:

Consider the ionic equation of this chemical equation. We are given barium chloride and silver nitrate as the reactants, and silver chloride and barium nitrate as the products. We can thus conclude that the ionic equation ( not balanced yet ) should be as follows -

Ba( 2 + ) + Cl ( - ) + Ag ( + ) + NO3 ( - ) ------> AgCl + Ba( 2 + ) + NO3( - )

As you can see these compounds are present in aqueous solutions, and are thus dissociated.

______________________________________________________

Now let us take a look at the number of elements on the reactant and product sides, and balance this chemical equation out -

Ba( 2 + ) + 2Cl ( - ) + 2Ag ( + ) + 2NO3 ( - ) ------> 2AgCl + Ba( 2 + ) + 2NO3( - )

Solution = Option C!

given that the only known ionic charges of lead are pb(ii) and pb(iv), how can you explain the existence of the pb2o3 salt

Answers

Answer:

See Explanation

Explanation:

Pb2O3 is better formulated as PbO.PbO2. It is actually a mixture of the two oxides of lead, lead II oxide and lead IV oxide.

This implies that this compound Pb2O3  (sometimes called lead sesquioxide) is a mixture of the oxides of lead in its two known oxidation states +II and +IV.

Hence Pb2O3  contains PbO and PbO2 units.