For the reaction 2CH₄ (g)+ 3 Cl₂ (g)→ 2 CHCl₃ (l)+ 3 H₂ (g),ΔH° = -118.6 kJ. ΔH°_f = -134.1 kJ/mol for CHCl₃ (l).Find ΔH°_f for CH₄ (g).

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The given reaction is: $2CH_4(g) + 3Cl_2(g) \rightarrow 2CHCl_3(l) + 3H_2(g)$, with $\Delta H^\circ = -118.6 \, \text{kJ}$.Given $\Delta H_f^\circ$ for $CHCl_3(l)$ is $-134.1 \, \text{kJ/mol}$.To find $\Delta H_f^\circ$ for $CH_4(g)$, we use the equation for the standard enthalpy change of the reaction ($\Delta H^\circ$) in terms of the standard enthalpies of formation ($\Delta H_f^\circ$) of the reactants and products:$$\Delta H^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})$$Substituting the given values:$$-118.6 = [2(-134.1) + 3(0)] - [2\Delta H_f^\circ(CH_4) + 3(0)]$$Solving for $\Delta H_f^\circ(CH_4)$:$$-118.6 = -268.2 - 2\Delta H_f^\circ(CH_4)$$$$2\Delta H_f^\circ(CH_4) = -268.2 + 118.6$$$$2\Delta H_f^\circ(CH_4) = -149.6$$$$\Delta H_f^\circ(CH_4) = -74.8 \, \text{kJ/mol}$$

For the Reaction 2CH4 (G)+ 3 Cl2 (G)→ 2 CHCl3

For the Reaction 2CH₄ (G)+ 3 Cl₂ (G)→ 2 CHCl₃