Question

when one mole of sodium bicarbonate is formed from its elements, 947.7 kj of energy is released into the surroundings. this enthalpy change can be represented as: na(s) + 1/2 h₂(g) + c(s) + 3/2 o₂(g) → nahco₃(s) + 947.7 kj. my answer, including all work: a na(s) + 1/2 h₂(g) + c(s) + 3/2 o₂(g) + 947.7 kj → nahco₃(s) b na⁺(aq) + hco₃⁻(aq) + 947.7 kj → nahco₃(s) c na⁺(aq) + hco₃⁻(aq) → nahco₃(s) + 947.7 kj

Explanation:

Step1: Identify the correct chemical - formation reaction

The formation of a compound from its elements should have the elements in their standard states on the reactant side. Sodium is in solid state (Na(s)), hydrogen is in di - atomic gas state ($H_2$(g)), carbon is in solid state (C(s)) and oxygen is in di - atomic gas state ($O_2$(g)). The reaction for the formation of sodium bicarbonate ($NaHCO_3$(s)) from its elements is $\text{Na}(s)+\frac{1}{2}\text{H}_2(g)+\text{C}(s)+\frac{3}{2}\text{O}_2(g)
ightarrow\text{NaHCO}_3(s)$. When 1 mole of $NaHCO_3$ is formed from its elements, 947.7 kJ of energy is released, so the correct reaction is $\text{Na}(s)+\frac{1}{2}\text{H}_2(g)+\text{C}(s)+\frac{3}{2}\text{O}_2(g)
ightarrow\text{NaHCO}_3(s)+947.7\text{ kJ}$.

Answer:

$\text{Na}(s)+\frac{1}{2}\text{H}_2(g)+\text{C}(s)+\frac{3}{2}\text{O}_2(g)
ightarrow\text{NaHCO}_3(s)+947.7\text{ kJ}$