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?

The reaction is said to be exothermic if the enthalpy change, ΔH, is negative and it's endothermic if ΔH is positive. In this case, the given ΔH is +252.8 kJ, which means that the reaction is endothermic. #b) Heat Transfer for 24.0 g CH3OH Decomposition#

First, we need to find the molar mass of CH3OH. The molecular formula is CH3OH, so the molar mass is: \(12.01 (C) + 3 \times 1.01 (H) + 16.00 (O) + 1.01 (H) = 32.04 \mathrm{~g/mol} \) Now, we can find the moles of CH3OH in 24.0 g: \(n = \frac{24.0 \mathrm{~g}}{32.04 \mathrm{~g/mol}}= 0.749 \mathrm{~mol} \)

From the balanced chemical equation, we can see that 2 moles of CH3OH decompose to release 252.8 kJ of heat. So, for each mole of CH3OH: \( q = \frac{252.8 \mathrm{~kJ}}{2} = 126.4 \mathrm{~kJ/mol} \) To find the amount of heat transferred during the reaction of 0.749 mol of CH3OH, we multiply q with the number of moles: \(q_{total} = 0.749 \mathrm{~mol} \times 126.4\mathrm{~kJ/mol} = 94.7 \mathrm{~kJ} \) # c) Methane Production for a given CH3OH Sample #

Given the enthalpy change is 82.1 kJ for this sample, we can calculate the moles of CH4 produced using the equation \(q = 126.4\mathrm{~kJ/mol} \times n \): \( n = \frac{82.1 \mathrm{~kJ}}{126.4 \mathrm{~kJ/mol}} = 0.6495 \mathrm{~mol} \)

Now, we can find the mass of CH4 by multiplying the moles with the molar mass of CH4, which is \( 12.01 (C) + 4 \times 1.01 (H) = 16.04 \mathrm{~g/mol} \): \(m = 0.6495 \mathrm{~mol} \times 16.04 \mathrm{~g/mol} = 10.4 \mathrm{~g} \) # d) Heat Released When 38.5 g of CH4 Reacts with O2 #

First, we need to find the moles of CH4 in 38.5 g: \(n = \frac{38.5 \mathrm{~g}}{16.04 \mathrm{~g/mol}}= 2.40 \mathrm{~mol} \)

As we calculated before, the reaction of 2 moles of CH3OH releases 252.8 kJ of heat. This means the reverse reaction, where 2 moles of CH4 react with O2 to form 2 moles of CH3OH, would absorb 252.8 kJ of heat. So, for each mole of CH4, the heat released is: \( q = - \frac{252.8 \mathrm{~kJ}}{2} = -126.4 \mathrm{~kJ/mol} \) To find the amount of heat released during the reaction of 2.40 mol of CH4, we multiply q with the number of moles: \(q_{total} = 2.40 \mathrm{~mol} \times (-126.4\mathrm{~kJ/mol}) = -303.4 \mathrm{~kJ} \) In conclusion: a) The reaction is endothermic. b) The amount of heat transferred for the decomposition of 24.0 g CH3OH is 94.7 kJ. c) When 82.1 kJ of enthalpy change in the reaction for a given sample of CH3OH, 10.4 g of methane gas is produced. d) The amount of heat released when 38.5 g of CH4 reacts with O2 to form CH3OH is -303.4 kJ.