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Ch.19 - Chemical Thermodynamics

Chapter 19, Problem 115

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is 2 H21g2 + O21g2 ¡ 2 H2O1l2 Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at 85 °C. (a) Use data in Appendix C to calculate ∆H° and ∆S° for the reaction. We will assume that these values do not change appreciably with temperature.

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Hey everyone, we're told that methane gas, the main component in bio gas is one of the clean fuels used for transportation. It is a useful fuel source. When it undergoes combustion. Using the data calculate the change in entropy and the change in entropy for the reaction assume that these values do not change significantly with temperature first. Let's go ahead and calculate our change in entropy. To calculate our change in entropy. We've learned that this is the standard entropy of formation of our products minus the standard entropy of formation of our reactant. Now let's go ahead and plug in our values. Starting off with carbon dioxide, We're going to multiply one mole By -393.5 kg joules per mole. And for each of these values we do have to put into account our moles. So for water we have to mole of water and we're going to multiply this by its Change in entropy of negative 285. Killer jewels Permal. Next we're going to subtract the standard entropy affirmation of our reactant. Starting off with our methane gas, we can see that we have one mole of methane and we're going to multiply this by negative 74.8 kg joules per mole. Next we're going to add our oxygen gas, we can see that we have to mull of oxygen from our reaction and we're going to multiply that by zero killer jules Permal. Now when we calculate this out and cancel out our units, We end up with a change in entropy of negative 890.4 kg jewels. Now let's go ahead and calculate our change in entropy for a change of entropy, we know that this is our entropy value of our products minus the entropy value of our reactant. Now let's go ahead and plug in these values again. Starting with carbon dioxide, we see we have one mole of carbon dioxide And we're going to multiply this by 213.6 jewels per mole kelvin. Next, we're going to add Two mol from our water and we're going to multiply two mole by 69. jules per mole Calvin. Now we're going to subtract our entropy values for our reactant. Starting with our methane gas, we see we have one mole and we're going to multiply this by 100 and 86.3 joules per mole kelvin. Next we're going to add two mol from our oxygen gas and multiply two mole by 205 point oh jules per mole kelvin. Now, when we calculate this out and cancel out all of our units, We end up with a change in entropy of negative 242.9 jewels per kelvin and this is going to be our final answers. Now I hope that made sense. And let us know if you have any questions