Consider the following reaction: $$ 2 \mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \quad \Delta H=+252.8 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(24.0 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is decomposed by this reaction at constant pressure. (c) For a given sample of \(\mathrm{CH}_{3} \mathrm{OH},\) the enthalpy change during the reaction is \(82.1 \mathrm{~kJ}\). How many grams of methane gas are produced? (d) How many kilojoules of heat are released when \(38.5 \mathrm{~g}\) of \(\mathrm{CH}_{4}(g)\) reacts completely with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CH}_{3} \mathrm{OH}(g)\) at constant pressure?

An endothermic reaction absorbs heat from the surroundings, making the enthalpy change, ΔH, a positive value. An exothermic reaction releases heat to the surroundings, making the enthalpy change, ΔH, a negative value. In this case, the enthalpy change is positive, \(ΔH = +252.8 kJ\). Therefore, this reaction is endothermic.

To calculate the heat transferred, we need to use the stoichiometry of the balanced equation and the molar mass of CH3OH to convert the mass given to moles. First, we need to find the molar mass of CH3OH. CH3OH has 1 carbon (C), 4 hydrogens (H), and 1 oxygen (O). Molar mass of CH3OH = \( 1 \times 12.01~g/mol (C) + 4 \times 1.008~g/mol (H) + 1 \times 16.0~g/mol (O) = 32.042~g/mol\) Now, let's find the number of moles of CH3OH. Number of moles of CH3OH = \(\dfrac{24.0~g~CH_{3}OH}{32.042~g/mol} = 0.749~mol\) From the balanced equation, 2 moles of CH3OH produce \(252.8~kJ\): 2 moles CH3OH → \(+252.8~kJ\) To calculate the heat transferred for 0.749 moles of CH3OH, we set up a proportion: \(0.749~mol~\times \dfrac{252.8~kJ}{2~mol} = Q\) Solve for Q to find the heat transferred: Q = \(0.749~mol \times 126.4~kJ/mol = 94.7~kJ\) Therefore, the heat transferred when 24.0 g of CH3OH is decomposed in this reaction at constant pressure is 94.7 kJ.

We are given that the enthalpy change for a sample of CH3OH is 82.1 kJ. We need to find the number of grams of CH4 produced in this case. First, let's find the number of moles of CH3OH. Number of moles of CH3OH = \(\dfrac{82.1~kJ}{126.4~kJ/mol} = 0.65~mol\) From the balanced equation, 2 moles of CH3OH produce 2 moles of CH4: 2 moles CH3OH → 2 moles CH4 So, the number of moles of CH4 produced = the number of moles of CH3OH (by stoichiometry) = \(0.65~mol\) Now, we need to find the mass of CH4 produced. The molar mass of CH4 is: 1 carbon (C) and 4 hydrogens (H): Molar mass of CH4 = \( 1 \times 12.01~g/mol (C) + 4 \times 1.008~g/mol (H) = 16.042~g/mol\) Mass of CH4 produced = \(0.65~mol \times 16.042~g/mol = 10.4~g\) Therefore, 10.4 g of methane gas is produced.

In this part, we are asked to find the heat released when 38.5 g of CH4 reacts completely with O2 to form CH3OH. This is essentially the reverse of the given reaction. We know that when a reaction is reversed, the sign of the enthalpy change is also reversed. So, for the reverse reaction, the enthalpy change, ΔH = -252.8 kJ Now, we need to find the number of moles of CH4 in 38.5 g: Moles of CH4 = \(\dfrac{38.5~g}{16.042~g/mol} = 2.40~mol\) From the balanced equation (in reverse), 2 moles of CH4 produce -252.8 kJ: 2 moles CH4 → \(-252.8~kJ\) To calculate the heat released for 2.40 moles of CH4, we set up a proportion: \(2.40~mol~\times \dfrac{-252.8~kJ}{2~mol} = Q\) Solve for Q to find the heat released: Q = \(2.40~mol \times -126.4~kJ/mol = -303~kJ\) Therefore, 303 kJ of heat is released when 38.5 g of CH4 reacts with O2 to form CH3OH at constant pressure. Note that the heat is released, so the value is negative.