Question

The ΔH°f of gaseous dimethyl ether (CH₃OCH₃) is -185.4 kJ/mol; the vapor pressure is 1.00 atm at -23.7°C and 0.526 atm at -37.8°C. (b) Calculate ΔH°f of liquid dimethyl ether.

Answer 1

The ΔH°f of liquid dimethyl ether is  + 975.204 kJ/mol.

The standard enthalpy of formation is a measure of the power launched or consumed when one mole of a substance is created below general situations from its pure elements. The symbol of the usual enthalpy of formation is ΔH f.

Enthalpies of formation assist us are expecting response enthalpies for plenty of reactions if the products and reactants are properly-studied, even though the precise response is new. To do this, we consider that we take the reactants and separate them into their natural elements in a widespread country.

ΔH product - ΔH reactant = Cp ( T₂ - T₁)

ΔH°f of liquid  -  ΔH°f of gaseous = 0.526 atm( -37.8°C) - ( -23.7°C)

ΔH°f of liquid = ΔH°f of gaseous *  0.526 * -10

                    =   -185.4 kJ/mol * -5.26    

                    = + 975.204 kJ/mol

Hence, ΔH°f of liquid dimethyl ether is  + 975.204 kJ/mol.

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