Video: Standard Free Energy Change for Hydrogenation of Ethene

Ethane gas can be produced by the hydrogenation of gaseous ethene. The standard entropies and enthalpies of formation for ethene and other materials are shown in the table. The standard change in Gibbs free energy, Δ𝐺_(298)^(⦵), for the hydrogenation of ethene at 298 K is expressed per mole of ethene reacted. Calculate, to 3 significant figures, the value of Δ𝐺_(298)^(⦵) at 298 K.

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Video Transcript

Ethane gas can be produced by the hydrogenation of gaseous ethene. The standard entropies and enthalpies of formation for ethene and other materials are shown in the table. The standard change in Gibbs free energy for the hydrogenation of ethene. At 298 Kelvin is expressed per mole of ethene reacted. Calculate, to three significant figures, the value of Δ𝐺 standard 298 at 298 Kelvin.

Ethane is one of the simplest hydrocarbons and one of the simplest alkanes. Hydrogenation is the addition of hydrogen to another substance. And ethene is the simplest alkene. We have the information we need to construct the reaction equation. Our first reactant is the gaseous ethene. Our second, hydrogen. And our product is gaseous ethane. All the substances in this reaction are gases. And the reaction is already balanced because we only need one equivalent of dihydrogen for each double bond.

The next thing I’m going to do is take a look at the data in the table. Standard molar entropy is the entropy per mole of a substance on a scale where a perfect crystal at zero degrees Kelvin has a molar entropy of zero. Here, the standard molar entropies have been given at 298 degrees Kelvin, which is indicated by the number in the bottom right. And the standard character in the top right indicates that all substances are in their standard states at one bar of pressure.

Meanwhile, the standard enthalpy of formation is the enthalpy change due to the formation of a substance from its elements per mole of substance with all substances in their standard states. For instance, the enthalpy of formation of ethene is the enthalpy change per mole of ethene when formed from carbon in its solid graphite form and hydrogen in its gaseous form.

The next step is to take the relevant information out of the table and put it next to the relevant chemical. That’s all the information transferred. Just be careful to ignore the data for ethyne and monatomic hydrogen gas. Now that we’ve got all the relevant information from the table, we can clear it away. The question has asked us to find the standard change in Gibbs free energy for the hydrogenation of ethene at 298 Kelvin.

Gibbs free energy is the maximum energy available from a system under constant temperature and pressure. It’s equivalent to the enthalpy of system minus the temperature times the entropy of the system. So, a change in Gibbs free energy is equal to the change in enthalpy minus the temperature multiplied by the change in entropy. So, we need to work out the change in enthalpy for this reaction and the change in entropy. The change in entropy is just equal to the final entropy, the entropy of the products, minus the initial entropy, the entropy of the reactants.

We want our final Δ𝐺 to be per mole of ethene. It’s fortunate that all the other components in this reaction are one to one with ethene. So, the combined entropy of our products is 229.2, the combined entropy of our reactants is 219.3 plus 130.7, and our units are joules per Kelvin per mole of ethene. So, our final value is minus 120.8 joules per Kelvin per mole of ethene. Now, we can move on to calculating the change in enthalpy.

The enthalpy change of a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. This equation arises from Hess’s law and the Hess’s cycle constructed by forming the reactants and products from their constituent elements. It arises because enthalpy is ultimately a state function, so it doesn’t matter which route you take from the reactants to the products as long as you get there.

So, per mole of ethene, the change in enthalpy for the formation of our products is minus 84.0 and for our reactants is 52.4 minus zero. It makes sense that the enthalpy of formation of H2 gas is zero because that’s already its standard state. And our units are kilojoules per mole of ethene. And the final value for enthalpy change is minus 136.4 kilojoules per mole of ethene. Before we plug these values into our Δ𝐺 expression, we should change the units of our entropy. The energy unit for our entropy change is joules, so we’ll convert that to kilojoules. We can do this by multiplying by one kilojoule per 1000 joules, giving us minus 0.1208 kilojoules per Kelvin per mole of ethene.

Let’s have a look at these numbers before we plug them in. The entropy change is negative, meaning that the reaction is exothermic. Meanwhile, there is a reduction in entropy of the reaction. This makes sense because we’re adding two gas molecules together. You could’ve used this fact to tell yourself that something was wrong if you got the opposite sign. Now, let moves on to the last part of this question, substituting our values into the equation.

So, the standard change in Gibbs free energy for the hydrogenation of ethene at 298 Kelvin is minus 136.4 minus 298 times negative 0.1208. And our units are kilojoules per mole of ethene. Remember, this equation only works when we use the temperature in Kelvin. Fortunately, that’s how it’s been provided. This all evaluates to minus 100.402 kilojoules per mole of ethene, with our final answer to three significant figures of minus 100 kilojoules per mole of ethene. This means that for each mole of ethene reacted with hydrogen, there is a reduction in the Gibbs free energy of the system by 100 kilojoules. Meaning that 100 kilojoules of energy is transmitted to the surroundings, likely as heat.

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