-4 Question 1: The standard formation enthalpies at 25.00 ° C for methanol (CH4O

Question

-4

Question 1:
The standard formation enthalpies at 25.00 ° C for methanol (CH4O (1)), water (H2O (1)) and carbon dioxide (CO2 (g)) are respectively -238.7 kJ mol, 285.8 kJ / mol, and -393.5 kJ / mol. Calculates the change in the environment (in J / K) when burning 13.5 g of methanol under a constant pressure of 1,000 atm at 25.00 oC (NB, combustion is the reaction of a substance with oxygen) Molecular to produce water and carbon dioxide).

Question 2:
At 25oC, for the 2A (aq) ----> B (aq) + C (aq) reaction, the equilibrium constant is 1.94. If the concentration of B (aq) was 0.365 M and the concentration of C (aq) was 0.470 M, what would be the minimum concentration of A (aq) to make this reaction spontaneous under these conditions?

Question 3:
For the reaction A (aq) + B (aq) <---> C (aq) + D (aq), the equilibrium constant is 23.8 at 25 ° C and 38.3 at 50 ° C What is the value of the variation in Gibbs standard free enthalpy (in kJ) of this reaction at 75oC?

Explanation / Answer

Ans 1

Methanol combustion reaction

CH3OH + 1.5O2 = CO2 + 2H2O

Enthalpy change for the reaction

= sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants

= 2*Hf(H2O) + Hf(CO2) - 1.5*Hf(O2) - Hf(CH3OH)

= 2*(-285.8) + (-393.5) - 0 - (-238.7)

= - 726.4 kJ/mol

Negative sign shows the heat release in environment

Moles of methanol = mass/molecular weight

= 13.5 g / 32g/mol

= 0.422 mol

Energy released into environment = 726.4 kJ/mol x 0.422 mol

= 306.45 kJ x 1000J/kJ

= 306450 J

change in the environment (in J / K)

= 306450 J / (25+273)K

= 1028.36 J/K

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