An electronic device requires two 1.50 V AA zine-carbon. Which, connected is ser

Question

An electronic device requires two 1.50 V AA zine-carbon. Which, connected is series, give 3.00 V. If the cell transfers two moles of electrons for each moles of reaction work can be performed by the transfers? 4.34 times 10^2 kJ 2.00 times 10^2 kJ 8.09 times 10^2 kJ 72.4 kJ 5.79 times 10^2 kJ Given the following data for the reaction A rightarrow B.determine the activation energy.E_infinity.of the reaction to three significant figures. 6.89 J/mol 57.3 J/mol 374 J/mol 39.9 J/mol 54.4 J/mol Which of the following is/are equal once equilibrium is established? the rates of the forward and reverse reactions the concentrations of reactant and products the time that a particular atom or molecule spends as a reactant and product reverse the rate constants of the forward and all of the above are equal Indicate which of the following has the lowest standard molar entropy H_2O(s) CH_3CH_2OH(t) CH_4(g) He(g) Na(s)

Explanation / Answer

5 (a) in translational motion the molecules completely moves from one position to other in a particular axis which requires highest amount of energy

6) when hydrogen reacts with metal to form hydride the oxidation state of

hydrogen = -1

Metal = +2

So answer : +2 and -1

7) Work = nFEcell

work = 2 X 96485 X 3 = 578910 Joules = 5.78 X 10^2 KJ

8) Activation energy can be calcualted by Arrhenius equation

lnK2 - lnK1 = Ea / R (1/T1- 1/T2)

K2 = 0.730 K1 =0.739

T2 = 250 K   T1 = 450 K

-0.315 - (-0.302) = Ea / 8.314 X 10^-3 ( 1/450 - 1/250)

Ea = 5.73 J / mole

9) at equilibrium

The rate of forward reaction = rate of backward reaction

Kf[reactants] = Kb [Products]

Kf/Kb= equilbirum constant = [Products] / [Reactants]

10) Lowest standard molar entropy will be of solid than liquid than gas

Now if number of atoms are more than entropy of molecule will be more

so answer : Na(s)

11) Given: Entropy = 161J/mol K

molecular weight of PbBr2 = 367 g / mole

So entropy of 367 grams of PbBr2 = 161 Joules

Entropy of 1 gram = 161 / 367 Joules

entropy of 2.45 grams = 161 x 2.45 / 367 = 1.07 J / K

-0.013 X 8.314 X 10^-3 / -0.00178 = Ea

Ea = 0.0607 KJ /mole =

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.