First slide

Hess law of constant summation

Question

Calculate ΔH^o for the reaction :
Na₂O + SO₃ → Na₂SO₄ given the following :
A) Na_(s) + H₂O_(l) → NaOH_(s) + 1/2 H_2(g) ; ΔH^o = – 146 kJ
B) Na₂SO_4(s) +H₂O→ 2NaOH_(s) + SO_3(g) ; ΔH^o = + 418 kJ
C) 2 Na₂O_(s) + 2H_2(g) → 4Na_(s) + 2H₂O_(l); ΔH^o = + 259 kJ

Moderate

Solution

$\begin{array}{l} 2 \times eq(A) \Rightarrow 2N{a_{(s)}} + {\rm{ 2}}{H_2}{O_{(l)}}{\rm{ }} \to {\rm{ 2}}NaO{H_{(s)}} + {\rm{ }}{H_{2(g)}};{\rm{ }}\Delta {H_1} = {\rm{ 2(}}-{\rm{ }}146{\rm{ )}}kJ\\\\ \,\,\,\,\,\frac{1}{{eq(B)}} \Rightarrow 2NaO{H_{(s)}} + {\rm{ }}S{O_{3(g)}} \to N{a_2}S{O_{4(s)}} + {H_2}{O_l}\,;\,\Delta {H_2} = - 418KJ\\\\ \underline {\frac{1}{2} \times eq.(C) \Rightarrow {\rm{ }}N{a_2}{O_{(s)}} + {\rm{ }}{H_{2(g)}} \to {\rm{ 2}}N{a_{(s)}} + {\rm{ }}{H_2}{O_{(l)}};\,\,\Delta {H_3} = \frac{{259}}{2}\,KJ\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ On\,addition:\,\,N{a_2}O{\rm{ }} + {\rm{ }}S{O_3} \to {\rm{ }}N{a_2}S{O_4}\,\,\,\,\,\,\,\Delta H = \,\,\,\Delta {H_1} + \,\,\Delta {H_2} + \,\,\Delta {H_3} \end{array}$

$\begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2( - 146) - 418 + \frac{{259}}{2}\,\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 580.5KJ\,\,\,\,\,\,\,\,\,\,\, \end{array}$

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