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Ch.5 - Thermochemistry

Chapter 5, Problem 110

We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethane: 2 CH4(g) → C2H6(g) + H2(g) Calculate the ΔH° for this reaction using the following thermochemical data: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH° = -890.3 kJ 2 H2(g) + O2(g) → 2 H2O(l) H° = -571.6 kJ 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) ΔH° = -3120.8 kJ

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Hi everyone here we have a question asking us to determine the entropy of the reaction. Using hess's law, silver oxide solid plus to hydrochloric acid. Aquarius forms to silver chloride solid plus hydrogen gas plus oxygen gas. So we're given the following data and let's go through each of these. So we have two silver solid plus half oxygen gasses forms silver oxide solid And are in the play is -31.1 kg per mole for this. To match our original equation, we need that reversed. So that's going to be 31.1 killed joules per mole. Next we have silver solid plus half chlorine. Gashes Forms silver chloride solid and the entropy is negative 127.0 killed joules per mole. And for that to match we need to multiply by two. So its entropy is going to be negative. 254 zero killed joules per mole. Next we have water liquid plus chlorine gasses forms to hydrochloric acid. Aquarius Plus half oxygen gasses and its entropy is negative 48.6 kg per mole. And we need that reversed. So that is 48 0.6 kill jules, Permal. And lastly we have hydrogen gasses plus oxygen gasses forms Water liquid and its entropy is negative 285.8 killed joules per mole. And we need that reversed. So that is 285 .8 killed joules per mole. So now let's look at what we have. We have silver oxide forms two silver plus half oxygen. We have two silver plus coring, forms two silver chloride. We have to hydrochloric acid plus one half oxygen forms water plus scoring. And we have water forms hydrogen and oxygen. So our two silver are going to cancel out. Our half oxygen are going to cancel out. Our chlorine are going to cancel out and our waters are going to cancel out. And that's going to leave us with silver oxide, solid plus to hydrochloric acid. A quiz forms two silver chloride plus hydrogen plus oxygen. So our entropy Is going to equal 31 .1 killed jules Permal plus Negative, zero Killer Jewels, thermal plus 48 0. kill jules per mole plus 0.8 kg joules per mole and that equals 111.5 kill jules Permal. And that is our final answer. Thank you for watching. Bye.
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Textbook Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C (b) Determine the amount of heat gained by the water. The specific heat of water is 4.184 J/1gK.

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Textbook Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C (d) What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

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