Formic acid, HCOOH, can be made by reacting methyl alcohol with oxygen gas. The

Question

Formic acid, HCOOH, can be made by reacting methyl alcohol with oxygen gas. The reaction is as follows: CH3OH (aq) + O2 (g) HCOOH (aq) + H2O (l) Given the following thermodynamic data: Substance Hof (kJ/mol)

So (kJ/mol K) O2 (g) 0.0 0.2050 H2O (l) -285.8 0.0699 CH3OH (aq) -246.0 0.1331 HCOOH (aq) -425.0 0.1630

a. Calculate the Ho (in kJ) and So (in kJ) for the process

. b. Explain the reason for the sign of the So.

c. Is this reaction spontaneous at 298 K? At 398 K? Explain both answers in terms of Go.

Explanation / Answer

a) dH rxn = dHf products - dHf reactants

dH rxn = {-425.0 - 285.8} - { 0 - 246.0}

dH rxn = -468.8 kJ/mol

dS rxn = So products - So reactants

dS rxn = {0.0699 + 0.1630} - { 0.2050 + 0.1331}

dSrxn = - 0.1052 kJ/mol

b) entrophy change decreases due to gaseous O2 became liquid H2O

order of entropy gas >> liquid > solid

c) dG = dH -TdS

at both T = 298 K and 398 K reaction is spontaneous because in both case we get dG values in negative

at T = 298 K

dG = -464.8 - 298 (-0.1052)

dG = -433.45 kJ/mol

at T = 398 K

dG = -464.8 - 398 (-0.1052)

dG = -422.93 kJ/mol

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