Consider the reaction: $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$ $\Delta H = -1.37 \times 10^3 kJ$ Consider the following propositions and decide which are true: I. The reaction is endothermic II. The reaction is exothermic. III. The enthalpy of the reaction would be different if the water formed was gaseous.
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The reaction is $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$ and $\Delta H = -1.37 \times 10^3 kJ$. Since $\Delta H$ is negative, the reaction is exothermic. Thus, statement II is true and statement I is false. Show more…
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C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -1.37 x 10^3 kJ/mol I. The reaction is exothermic. II. The enthalpy change would be different if gaseous water were produced. III. The reaction is not an oxidation-reduction reaction.
David C.
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -1.37x10^3 kJ What would happen to the value of ΔH for the reaction of ethanol with oxygen if each of the following occurred: If the reaction was reversed: 2CO2(g) + 3H2O(l) → C2H5OH(l) + 3O2(g) ΔH = 1.37x10^3 kJ If the reaction was multiplied by 2: 2C2H5OH(l) + 6O2(g) → 4CO2(g) + 6H2O(l) ΔH = -2.74x10^3 kJ If the water was (g) instead of (l): C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) ΔH = -1.37x10^3 kJ If the ethanol was (g) instead of (l): C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -1.37x10^3 kJ Is this reaction exothermic or endothermic: This reaction is exothermic.
Adi S.
The cnthalpy change for the formation of $\mathrm{II}_{2} \mathrm{O}(1)$ from its elements in their standard state is $572 \mathrm{~kJ}$; therefore, (1) the reaction is cndothermic (2) the sum of the enthalpy of lydrogen and that of oxygen must be greater than that of water(3) the sum of the enthalpy of hydrogen and that of oxygen must be less than that of water (4) the enthalpy of water must be zero
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