Methanol 1CH3OH2 is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane 1CH42: Step 1. CO1g2 + 2 H21g2 S CH3OH1l2 ΔS° = - 332 JK Step 2. CH3OH1l2 S CH41g2 + 12 O21g2 ΔS° = 162 JK (a) Calculate ΔH° in kilojoules for step 1.

Question

Methanol 1CH3OH2 is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane 1CH42: Step 1. CO1g2 + 2 H21g2 S CH3OH1l2 ΔS° = - 332 J>K Step 2. CH3OH1l2 S CH41g2 + 1>2 O21g2 ΔS° = 162 J>K (a) Calculate ΔH° in kilojoules for step 1.

Accepted Answer

Hello everyone today. We are being given the following problem. In two steps, ethanol is produced industrially from carbon monoxide and hydrogen gas. It is being considered for use as a precursor to hydrocarbon fuels like methane because it's inexpensive to produce. And then we have step one and step two here. Our question is, what is the entropy of the reaction in killing jewels for step one? So essentially we're gonna be focusing on this first step right here. And the first we want to do is they want to note our entropy of formations for various compounds in our step one. So we want it for our carbon monoxide gas. We want the entropy of formation for our ethanol. We also want our entropy a formation for our hydrogen gas and we want the entropy of formation for our water. First the entropy of formation for our hydrogen gas is zero since it is a mono atomic ion or a monotonic adam for our carbon monoxide is negative 110.5 kg joules per mole for our ethanol. It is negative 277.6 kg joules per mole. And lastly, for our water it's going to be negative 285.8 kg joules per mole. Now that we have these values out of In order we can go ahead and set up our equation. So our entropy a formation for the reaction is going to be equal to our entropy of formation of our products minus the entropy of formation for our reactant. So for our products we have our ethanol and we have our water. And so we set the heat of formation or the entropy of formation for our ethanol was negative 277. kg joules per mole. And we're going to add that to the entropy of formation of our water, which is negative 2 85.8 kg joules per mole. We're going to take that number. That is the resulting number. We're going to subtract that by the heat of formation of our product of our reactant, which is going to be our ceo to our carbon monoxide and hydrogen gas, which we said our hydrogen gas was zero. So we only need to take into account our two moles of our carbon monoxide, which has a heat of formation of negative 100.5 kg joules per mole. So why did I multiply by 2? Well, the coefficient in front of that carbon monoxide is two. So that tells us that we have two moles of that. When we simplify this equation, we get the heat of formation for our reaction is negative 342.4 kg joules per mole. As our final answer. Overall, I hope that this helped and until next time

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Key Concepts

Enthalpy Change (ΔH°)

Gibbs Free Energy and Entropy (ΔS°)

Stoichiometry in Chemical Reactions