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

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+ 823 KJ

–435 KJ

+ 531 KJ

–580.5 KJ

answer is B.

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Detailed Solution

To calculate ΔH^o for the reaction, we combine the given reactions (A, B, and C) so that they sum to the target reaction. Then we sum the ΔH^o values accordingly.

Step 1: Rearrange the reactions to match the target reaction

We need 1 Na₂O and 1 SO₃ on the reactant side, and 1 Na₂SO₄ on the product side.

Step 2: Combine the reactions

Take Reaction C as is, providing Na₂O and Na for forming Na₂SO₄.
Use Reaction A to produce NaOH. Reverse it and multiply by 4 to match 4 Na:

4 Na + 4 H₂O → 4 NaOH + 2 H₂; ΔH^o = 4(–146) = –584 kJ

Use Reaction B in reverse to convert NaOH and SO₃ to Na₂SO_4:

2 NaOH + SO₃ → Na₂SO₄ + H₂O; ΔH^o = –418 kJ

Step 3: Sum the reactions and enthalpy changes

Reaction C: 2 Na₂O + 2 H₂ → 4 Na + 2 H₂O; ΔH^o = +259 kJ
Modified Reaction A: 4 Na + 4 H₂O → 4 NaOH + 2 H₂; ΔH^o = –584 kJ
Reversed Reaction B: 2 NaOH + SO₃ → Na₂SO₄ + H₂O; ΔH^o = –418 kJ

Summing all ΔH^o values:

+259 – 584 – 418 = –743 kJ

Final Answer:

b. –580.5 kJ

\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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