Respuesta :
Answer:
A) The ΔU° of [tex]K_2O[/tex] is -361 kJ/mol.
B) The ΔH° of [tex]K_2O[/tex] is -362 kJ/mol.
Explanation:
Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water
[tex]q=[q_1+q_2][/tex]
[tex]q=[c_1\times \Delta T+m_2\times c_2\times \Delta T][/tex]
where,
q = heat released by the reaction
= heat absorbed by the calorimeter
[tex]q_2[/tex] = heat absorbed by the water
[tex]c_1[/tex] = specific heat of calorimeter = [tex]1849 J/K[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/gK[/tex]
[tex]m_2[/tex] = mass of water = 1439 g
[tex]\Delta T[/tex] = change in temperature = [tex]1.60 K[/tex]
Now put all the given values in the above formula, we get:
[tex]q=[(1849 J/K \times 1.60 K)+(1439 g \times 4.184J/gK\times 1.60 K)][/tex]
[tex]q= 12,591.64 J[/tex]
[tex]4K(s)+O_2(g)\rightarrow 2K_2O(g)[/tex]
[tex]2K(s)+\frac{1}{2}O_2(g)\rightarrow K_2O(g)[/tex]
m = mass of substance= 2.50 g
n = moles of substance
[tex]n=\frac{2.720 g}{39 g/mol}=0.06974 mol[/tex]
According to reaction , 2 moles of K gives 1 mole of [tex]K_2O[/tex]
Then 0.06974 moles of K will give:
[tex]\frac{1}{2}\times 0.06974 mol=0.03487 mol[/tex] of [tex]K_2O[/tex]
12,591.64 Joules of energy was released when 0.03487 mol of [tex]K_2O[/tex] are formed.
So, the enthaply of formation of [tex]K_2O[/tex]: ΔU°
[tex]=\frac{-12,591.64 J}{0.03487 mol}=-361,102.38 J/mol=-361.10 kJ/mol[/tex]
The ΔU° of [tex]K_2O[/tex] is -361 kJ/mol.
[tex]\Delta H^o =\Delta U^o +\Delta n_gRT[/tex]
[tex]\Delta n_g[/tex] = moles of gases on RHS - moles of gasses on LHS
[tex] = -361,102.38 J/mol +(0-\frac{1}{2})\times 8.314 J/mol\times 298.15 K[/tex]
[tex]\Delta H^o=-362 kJ/mol[/tex]
The ΔH° of [tex]K_2O[/tex] is -362 kJ/mol.
