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 a. na(s) + 1/2 h₂(g) + c(s) + 3/2 o₂(g) → nahco₃(s) + 947.7 kj b. na(s) + 1/2 h₂(g) + c(s) + 3/2 o₂(g) → nahco₃(s) - 947.7 kj c. na⁺(aq) + hco₃⁻(aq) → nahco₃(s) + 947.7 kj d. na⁺(aq) + hco₃⁻(aq) → nahco₃(s) - 947.7 kj my answer, including all work:

Explanation:

Step1: Identify the reaction of formation of sodium bicarbonate

The formation of sodium bicarbonate from its elements involves the combination of sodium (Na), hydrogen (H₂), carbon (C) and oxygen (O₂) to form NaHCO₃. The enthalpy change of formation is the energy released when one - mole of the compound is formed from its elements in their standard states.

Step2: Analyze the given equations

The correct equation for the formation of sodium bicarbonate from its elements should have the elements on the left - hand side in their standard states and sodium bicarbonate on the right - hand side. The energy released is represented as a positive value on the left - hand side if the reaction is exothermic (which is the case here as energy is released). The standard states of Na is solid (s), H₂ is gas (g), C is solid (s) and O₂ is gas (g).
The balanced equation for the formation of one mole of NaHCO₃(s) from its elements is:
\[Na(s)+\frac{1}{2}H_{2}(g)+C(s)+\frac{3}{2}O_{2}(g)
ightarrow NaHCO_{3}(s)\]
The energy released is 947.7 kJ, so the correct representation is \(Na(s)+\frac{1}{2}H_{2}(g)+C(s)+\frac{3}{2}O_{2}(g)
ightarrow NaHCO_{3}(s)+947.7\ kJ\)

Answer:

The correct chemical equation representing the formation of one mole of sodium bicarbonate from its elements with the release of 947.7 kJ of energy is \(Na(s)+\frac{1}{2}H_{2}(g)+C(s)+\frac{3}{2}O_{2}(g)
ightarrow NaHCO_{3}(s)+947.7\ kJ\)