Question

Calculate \(\Delta H_f^\circ\) for the reaction:
\[ Na_2O(s) + SO_3(g) \rightarrow Na_2SO_4(g) \]
given the following:
(A) \(Na(s)+H_2O(l)\rightarrow NaOH(s)+\frac{1}{2}H_2(g)\), \(\Delta H^\circ=-146\,kJ\)
(B) \(Na_2SO_4(s)+H_2O(l)\rightarrow 2NaOH(s)+SO_3(g)\), \(\Delta H^\circ=+418\,kJ\)
(C) \(2Na_2O(s)+2H_2(g)\rightarrow 4Na(s)+2H_2O(l)\), \(\Delta H^\circ=+259\,kJ\)

• A. \(+823\,kJ\)
• B. \(-581\,kJ\)
• C. \(-435\,kJ\)
• D. \(+531\,kJ\)

Hint:

In Hess law problems, reverse equations to place products/reactants properly, then add so that unwanted species cancel to get the target reaction.

Answer 1

The Correct Option is B

Step 1: Use Hess’s law.
We need:
\[ Na_2O + SO_3 \rightarrow Na_2SO_4 \]
Step 2: Manipulate given equations.
(A) Multiply by 2:
\[ 2Na + 2H_2O \rightarrow 2NaOH + H_2,\quad \Delta H=-292 \]
(C) Reverse and divide by 2:
\[ 2Na + H_2O \rightarrow Na_2O + H_2,\quad \Delta H=-129.5 \]
(B) Reverse:
\[ 2NaOH + SO_3 \rightarrow Na_2SO_4 + H_2O,\quad \Delta H=-418 \]
Step 3: Add appropriately to cancel intermediates.
After summation and cancellation, net becomes:
\[ Na_2O + SO_3 \rightarrow Na_2SO_4 \]
Total enthalpy becomes:
\[ \Delta H = -581\,kJ \]
Final Answer:
\[ \boxed{-581\,kJ} \]