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What is the change in enthalpy in joules when 5.44 x 10- mol of AgCl dissolves in water according to the following chemical equation:
AgCl(s) rightarrow Ag+(aq)+ Cl-(aq) AH=65.5 kj

Respuesta :

sebassandin

Answer:

[tex]\Delta H=35.3J[/tex]

Explanation:

Hello,

In this case, since the dissolution of silver chloride involves a change in enthalpy of 65.5kJ per 1 mole  when undergone, for 5.44x10⁻⁴ moles, the enthalpy change is:

[tex]\Delta H=65.5\frac{kJ}{1mol}*5.44x10^{-4}mol\\ \\\Delta H=0.0353kJ*\frac{1000J}{1kJ}\\ \\\Delta H=35.3J[/tex]

Best regards.

Lanuel

The change in enthalpy (in joules) when AgCl dissolves in water is 35.362 Joules.

Given the following data:

  • Number of moles ([tex]A_gC_l_{(s)}[/tex] ) = [tex]5.44[/tex] × [tex]10^{-4}[/tex] mol.

To find the change in enthalpy (in joules) when AgCl dissolves in water according to the following chemical equation:

                                 [tex]A_gC_l_{(s)}[/tex]  -----> [tex]Ag^{+}_{(aq)} + Cl^{-}_{(aq)}[/tex]    [tex]\Delta H = 65.5 \;kJ/mol[/tex]

1 mole of silver chloride =  [tex]65.5[/tex]  × [tex]10^{3}[/tex] J/mol

[tex]5.44[/tex] × [tex]10^{-4}[/tex] mole of silver chloride = X J/mol

Cross-multiplying, we have:

[tex]X =[/tex] [tex]5.44[/tex] × [tex]10^{-4}[/tex] × [tex]65.5[/tex]  × [tex]10^{3}[/tex]

[tex]X = 0.544[/tex] × [tex]65.5[/tex]

X = 35.362 Joules.

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