**Background Info:**
The standard enthalpy of formation \((\Delta H^\circ_f)\) is the enthalpy change that occurs when exactly 1 mol of a compound is formed from its constituent elements under standard conditions. The standard conditions are 1 atm pressure, a temperature of 25 °C, and all species present at a concentration of 1 M.
A "standard enthalpies of formation table" might look something like this:
| Substance | \(\Delta H^\circ_f\) (kJ/mol) |
|-----------|-------------------------------|
| H(g) | 218 |
| H\(_2\)(g)| 0 |
| Ba(s) | 0 |
| Ba\(^{2+}\)(aq) | −538.4 |
| C(g) | 71 |
| C(s) | 0 |
| N(g) | 473 |
| O\(_2\)(g)| 0 |
| O(g) | 249 |
| S\(_2\)(g)| 129 |
**Question:**
What is the balanced chemical equation for the reaction used to calculate \(\Delta H^\circ_f\) of BaCO\(_3\)(s)?
If fractional coefficients are required, enter them as a fraction (i.e. 1/3). Indicate the physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for aqueous solution.
**EXPRESS ANSWER AS A CHEMICAL EQUATION.**
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Answer (3)
You will get at first carbon dioxide by burning C: C + O 2 = C O 2 Burn hydrogen to obtain water: H 2 + 2 1 O 2 = H 2 O Combine them: C O 2 + H 2 O = H 2 C O 3 Now react it with Ba: B a + H 2 C O 3 = B a C O 3 + H 2 To sum up, the reaction is B a + C + 2 3 O 2 = B a C O 3 , using hydrogen as a catalyst.
You will get at first carbon dioxide by burning C:
C+O_2=CO_2
Burn hydrogen to obtain water:
H_2+\frac{1}{2}O_2=H_2O
Combine them:
CO_2+H_2O=H_2CO_3
Now react it with Ba:
Ba+H_2CO_3=BaCO_3+H_2
To sum up, the reaction is Ba+C+\frac{3}{2}O_2=BaCO_3, using hydrogen as a catalyst. ;
The balanced chemical equation for the formation of BaCO₃(s) is Ba(s) + C(s) + \frac{3}{2} O₂(g) -> BaCO₃(s). This equation illustrates the combination of barium, carbon, and oxygen in their standard states. The reaction yields one mole of barium carbonate as a solid product.
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