Answer:
The true statements are: I. The reaction is exothermic.
II. The enthalpy change would be different if gaseous water was produced.
Explanation:
The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ
1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .
Since the enthalpy change for a combustion reaction is negative. <u>Therefore, the given reaction is exothermic.</u>
2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.
ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)
As the value of ΔHf° of water in gaseous state and liquid state is not the same.
<u>Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.</u>
3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.
The given chemical reaction, represents the combustion reaction of ethanol.
Since combustion reactions are redox reactions. <u>Therefore, the given combustion reaction is an oxidation-reduction reaction.</u>
4. According to the ideal gas equation: P.V =n.R.T
Volume (V) ∝ n (number of moles of gas)
Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.
<u>Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.</u>
![pH=-log[H^{+}]=-log[H_{3}O]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D%3D-log%5BH_%7B3%7DO%5D%20%20)
. Using the value in formula, we get:![pH=-log[1.8*10^{-4}]=3.745](https://tex.z-dn.net/?f=pH%3D-log%5B1.8%2A10%5E%7B-4%7D%5D%3D3.745%20)