FeCl_3 can be prepared from iron and chlorine in two ways: Which is true of the

Question

FeCl_3 can be prepared from iron and chlorine in two ways: Which is true of the enthalpy changes of these routes? The single-step route has an enthalpy change that can be larger or smaller than the two-step route, depending upon conditions. The single-step route will have a smaller enthalpy change than the two-step route. The single-step route will have a larger enthalpy change than the two-step route. The overall enthalpy changes for the two routes are the same. Use these data to determine the enthalpy change for this reaction. Na_2 O_2 (s) rightarrow Na_2 O(s) + 1/2 O_2(g) +99 kJ -99 kJ +198 kJ cannot determine without deltaH_f for O_2(g) Chlorine will oxidize I^- as shown in this equation: Cl_2 + 2I^- rightarrow I_2 + 2Cl^- A buried metal tank will not corrode if it is wired to a metal rod that corrodes more easily. Which combination protects the tank from corrosion? Mg rod and Sn tank Fe rod and Zn tank Fe rod and Al tank Cu rod and Zn tank

Explanation / Answer

Q15

FeCl3

For the enthalpy change roues:

a)

False, the final change should be the SAME

b)

false, it will have the SAME total enthalpy change

c)

False, the change must be the same

d)

This is true, the overall enthalpy change should be the SAME

Q16

HRxn = Hproducts - Hreactants

HRxn = H-Na2O + 1/2O2 - (Na2O2)

HRxn = -416) + 1/2*0 - (-515) = 99 kJ/mol

choose A

note that for Hf of O2, it is 0 since it is elemental state

Q17

Br2 from Br-:

Recall that reactivity goes as follows:

F2 > Cl2 > Br2 > I2

so

The only specie that will work for Bromine is also I-

so

Br2 + 2I- = 2 + 2Br-

which is what we wanted

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