when one mole of sodium bicarbonate is formed from its elements, 947.7 kj of energy is released into the…

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

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

Answer

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 $\text{Na}(s)+\frac{1}{2}\text{H}_2(g)+\text{C}(s)+\frac{3}{2}\text{O}_2(g)\rightarrow\text{NaHCO}_3(s)$. When 1 mole of $NaHCO_3$ is formed from its elements, 947.7 kJ of energy is released, so the correct reaction is $\text{Na}(s)+\frac{1}{2}\text{H}_2(g)+\text{C}(s)+\frac{3}{2}\text{O}_2(g)\rightarrow\text{NaHCO}_3(s)+947.7\text{ kJ}$.

Answer:

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