Given the following data: C2 H2(g)+(5)/(2) O2(g) →2 CO2(g)+H2 O(l)ΔH=-1300 . kJ C(s)+O2(g) →CO

step 1 To get delta H for this reaction, we must take a few steps. First, we have this reaction. It was

Given the following data: C2 H2(g)+(5)/(2) O2(g) →2...

Given the following data: $$\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\Delta H=-1300 . \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \quad \rightarrow \mathrm{CO}_{2}(g) \Delta H=-394 \mathrm{kJ}$$ $$\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \quad \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H=-286 \mathrm{kJ}$$ calculate $\Delta H$ for the reaction $$2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)$$

Given the following data:
$$\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\Delta H=-1300 . \mathrm{kJ}$$
$$\mathrm{C}(s)+\mathrm{O}_{2}(g) \quad \rightarrow \mathrm{CO}_{2}(g) \Delta H=-394 \mathrm{kJ}$$
$$\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \quad \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H=-286 \mathrm{kJ}$$
calculate $\Delta H$ for the reaction
$$2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)$$

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Key Concepts

Hess's Law

Hess's Law states that the total enthalpy change of a reaction is the same regardless of the pathway taken, since enthalpy is a state function. This principle allows one to combine or rearrange different reactions with known enthalpy changes to calculate the enthalpy change of an overall reaction.

Enthalpy Change (?H)

The enthalpy change, ?H, is a measure of the heat absorbed or released during a chemical reaction at constant pressure. It is an essential thermodynamic quantity that indicates whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).

Standard Enthalpy of Formation

The standard enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states. In these calculations, elements in their most stable form at standard conditions (such as C(s) and H?(g)) have an enthalpy of formation of zero.

Reaction Manipulation Techniques

Reaction manipulation techniques involve algebraically adding, subtracting, reversing, or scaling chemical equations to achieve a desired overall reaction. Each manipulation affects the associated enthalpy changes, such that reversing a reaction inverts the sign of ?H and multiplying a reaction by a coefficient multiplies the enthalpy change by the same factor.

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