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abruzzese [7]
3 years ago
15

The standard internal energy change for a reaction can be symbolized as Δ U ∘ rxn or Δ E ∘ rxn . For each reaction equation, cal

culate the energy change of the reaction at 25 ∘ C and 1.00 bar . Sn ( s ) + 2 Cl 2 ( g ) ⟶ SnCl 4 ( l ) Δ H ∘ rxn = − 511.3 kJ/mol
Chemistry
1 answer:
rosijanka [135]3 years ago
6 0

Answer : The internal energy change is, -506.3 kJ/mol

Explanation :

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -511.3kJ/mol=-511300.0J/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles

Change in moles = Number of moles of product side - Number of moles of reactant side

According to the reaction:

Change in moles = 0 - 2 = -2 mole

That means, value of \Delta n_gRT = 0

R = gas constant = 8.314 J/mol.K

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above formula, we get

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-511300.0J/mol)-[-2mol\times 8.314J/mol.K\times 298K

\Delta U=-511300.0J/mol+4955.144J/mol

\Delta U=-506344.856J/mol=-506.3kJ/mol

Therefore, the internal energy change is -506.3 kJ/mol

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When I2 and FeCl2 are mixed together, iodine (I) cannot replace chlorine (Cl) in the compound because iodine is lower on the per
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Answer: The given statement is true.

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has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a
Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

               A + B \rightarrow C

Initial :     0.3   0.4          0

Change:  -x       -x           x

Equilbm: (0.3 - x)  (0.4 - x)  x  

We know that, relation between standard free energy and equilibrium constant is as follows.

      \Delta G = -RT ln K

Putting the given values into the above formula as follows.

      \Delta G = -RT ln K

      -3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})

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  •  [B] = 0.4 - 0.1417

       = 0.2583 M

  •  [C] = 0.1417 M
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