Question

The molar heat capacity of silver is 25.35 J/mol⋅∘C. How much energy would it take to raise the temperature of 9.10 g of silver by 10.5 ∘C? Use the following values:specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C)

Find ΔHrxn for the reaction: N2O(g)+NO2(g)→3NO(g) Use these reactions with known ΔH's: 2NO(g)+O2→2NO2(g)ΔH=−113.1kJ N2(g)+O2→2NO(g)ΔH=+182.6kJ 2N2O(g)→2N2(g)+O2(g)ΔH=−163.2kJ

Answer 1

Given the molar heat capacity of silver = 25.35 J/moloC

We know that the molar mass of silver = 108 g/mol

So the specific heat capacity of silver,c = molar heat capacity of silver / molar mass

                                                        = 25.35 / 108

                                                        = 0.235 J/goC

The amount of heat required to raise its temperature is , Q = mcdt

Where m = mass of silver = 9.10 g

dt = raise of temperature = 10.5 oC

PLug the values we get Q = 9.10x0.235 x 10.5 J

                                      = 22.43 J

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N2O(g)+NO2(g)→3NO(g)   ; ΔH_rxn = ?       ----(1)

2NO(g)+O2→2NO2(g)   : ΔH₁ =−113.1kJ   ----(2)

N2(g)+O2→2NO(g)     : ΔH₂ =+182.6kJ      ----(3)

2N2O(g)→2N2(g)+O2(g) : ΔH₃ =−163.2kJ      -----(4)

Eqn(1) can be obtained from the remaining equations as follows :

Eqn(1) = [(1/2)xreverse of Eqn(1)]+Eqn(3) + [(1/2)xEqn(4)]

So ΔH_rxn = [(1/2) x(- ΔH₁ )]+ ΔH₂ + [(1/2)x ΔH₃ ]

              = [(1/2x(-(-113.1))] + (+182.6) + [(1/2)x(-163.2)]

             = +157.55 kJ