Question

Calculate $\Delta H^o$ for the reaction,
$N{a}_{2}O\left(s\right)+S{O}_3\left(g\right)\to N{a}_{2}S{O}_4\left(g\right)$ using the following :
$Na\left(s\right)+H_{2}O\left(l\right)\to NaOH\left(s\right)+\frac{1}{2}{H}_2\left(g\right);\Delta H^o=-146kJ$
$N{a}_{2}S{O}_4\left(s\right)+H_{2}O\left(l\right)\to 2NaOH\left(s\right)+S{O}_3\left(g\right);\Delta H^o=+418kJ$
$2NaO\left(s\right)+2H_2\left(g\right)\to 4Na\left(s\right)+2H_{2}O\left(l\right);\Delta H^o=+259kJ$

• A. + 823 kJ
• B. - 581 kJ
• C. - 435 kJ
• D. + 531 kJ

Answer 1

Answer: (B)

- 581 kJ

$N{a}_{\left(s\right)}+{H_{2}O}_{\left(l\right)}\to NaO{H}_{\left(s\right)}+1/2{H_2}_{\left(g\right)}$ ; $\Delta H_1^o=-146kJ⟶\left(1\right)$

${N{a}_{2}S{O}_4}_{\left(s\right)}+{H_{2}O}_{\left(l\right)}\to 2NaO{H}_{\left(s\right)}+{S{O}_3}_{\left(g\right)}$ ; $\Delta H_2^o=+418kJ⟶\left(2\right)$

${2N{a}_{2}O}_{\left(s\right)}+{2H_2}_{\left(g\right)}\to 4N{a}_{\left(s\right)}+{2H_{2}O}_{\left(l\right)}$ ; $\Delta H_3^o=+259kJ⟶\left(3\right)$

${N{a}_{2}O}_{\left(s\right)}+{S{O}_3}_{\left(g\right)}\to {N{a}_{2}S{O}_4}_{\left(g\right)}⟶\left(4\right)$

The required reaction equation can be obtained by $(equation3÷2)+(equation1\times 2)-\left(equation2\right)$

$\Rightarrow \Delta H^o=\frac{+259}{2}+(-146)\times 2-(+418)=-580.5\approx -581kJ$ .