Asked by Anonymous
Write the balanced chemical reaction (showing appropriate symbols and states) for the chemical reaction with enthalpy change equal to ΔHf° [NH3(g)]
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Answered by
Side Walks
Calculate the enthalpy for this reaction:
2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ
Given the following thermochemical equations:
C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1299.5 kJ
C(s) + O2(g) ---> CO2(g) ΔH° = -393.5 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ
2CO2(g) + H2O(ℓ) ---> C2H2(g) + (5/2)O2(g) ΔH° = +1299.5 kJ
2C(s) + 2O2(g) ---> 2CO2(g) ΔH° = -787 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ
+1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ Good Luck
2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ
Given the following thermochemical equations:
C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1299.5 kJ
C(s) + O2(g) ---> CO2(g) ΔH° = -393.5 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ
2CO2(g) + H2O(ℓ) ---> C2H2(g) + (5/2)O2(g) ΔH° = +1299.5 kJ
2C(s) + 2O2(g) ---> 2CO2(g) ΔH° = -787 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ
+1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ Good Luck
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