CH4(g) + Cl2(g)  CH2Cl2(g) + 2 HCl(g) Use the data in the table below to...

CH4(g) + Cl2(g)  CH2Cl2(g) + 2 HCl(g) Use the data in the table below to calculate the standard enthalpy, ∆H˚, for the reaction above. Substance CH4(g) CH2Cl2(g) HCl(g) ∆Hf˚, kJ•mol-1 –74.6 –95.4 –92.3

Expert Solution

_fH⁰ CH_4(g) = -74.6 KJ/mol

_fH⁰ Cl_2(g) = 0 KJ/mol

_fH⁰ CH₂Cl₂(g) = -95.4/mol

_fH⁰ HCl_(g) = -92.3 KJ/mol

H⁰ = _fH⁰ (Product) - _fH⁰ (Reactant)

H⁰ = [(_fH⁰ CH₂Cl₂(g) ) + (2_fH⁰ HCl_(g))] - [(_fH⁰ CH_4(g) ) + (2_fH⁰ Cl₂)]

= [ (-95.4) + (2 (-92.3)] - [(-74.6) + (0)]

= [-280] - [-74.6]

H⁰ = -205.4 KJ/mol

H⁰ for reaction = -205.4 KJ/mol

queen_honey_blossom answered 1 year ago

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