Ethyl chloride 1C2H5Cl2, a substance used as a topical anes-thetic, is prepared by reaction of ethylene with hydrogen chloride:
C2H41g2 + HCl1g2 ¡ C2H5Cl1g2 ΔH° = - 72.3 kJ
How much PV work is done in kilojoules, and what is the value of ΔE in kilojoules if 89.5 g of ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is - 71.5 L?

Question

Ethyl chloride 1C2H5Cl2, a substance used as a topical anes-thetic, is prepared by reaction of ethylene with hydrogen chloride: 
C2H41g2 + HCl1g2 ¡ C2H5Cl1g2 ΔH° = - 72.3 kJ 
How much PV work is done in kilojoules, and what is the value of ΔE in kilojoules if 89.5 g of ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is - 71.5 L?

Accepted Answer

Hello everyone today. We have the following problem. Consider the reaction of nitric oxide with oxygen gas to form nitrogen dioxide. And then we have the following formula here, calculate the work done in killing jewels and calculate the change in energy and kill jules. When 70.2 g of nitric oxide reacts with 81.6 g of oxygen gas at a consistent pressure of 1.2 atmospheres and a volume change of negative 62.5 liters. So first we have to recall the formula for work. The formula for work is that work is equal to our negative pressure times change in our volume. Now we were given both these variables in the problem So we're gonna just plug them in for atmospheres we're going to do 1.20 atmospheres Times are changing volume which is -62.5 L. We're going to get 75 Leaders Times atmospheres. However, we need to convert this and get rid of our atmospheres so that we're just left with our killer jewels. And what we can do with this is we can multiply by the conversion factor that There's 1 01.325 jewels for every Leader Times atmosphere. And when we do that we get 7,599.375. And then her jewels because our units for Leaders atmosphere is canceled out. And then we multiply the conversion factor that one kg is equal to 10 to the third jewels. Units of joules cancel out and we were left with work equally 7.0 or 7.60 kg jewels. So that's the first part of our problem. Next, we need to determine what our limiting reactant is. So we need to find the limiting reactant. Now we saw for our moles of nitric oxide and nitrogen oxide and oxygen. So for our moles of Nitric oxide, nitric oxide, we take however many we have, which was 70-70.2 g. And we multiply that by the Molar Mass, which is going to be really close to 30 g. And then we have one mole And once our units of g canceled out, we're gonna be left with 2.34 moles of nitric oxide. We then find our moles of our di nitrogen nitrogen dioxide. We do that. We take the Molds that we just solved for which was our 2. moles of our nitric oxide. What we can do is we can use our multiple ratio that for every two moles Of our nitrogen dioxide, we have two moles of nitric oxide. And those numbers came from our formula here. The coefficients there are moles of nitric oxide canceled out. We're left with 2.4 moles of nitrogen dioxide. So essentially we could have just multiplied by one since 2/2 is equal to one. Next we can find our moles of oxygen gas. We're going to take however many we had, which was 81.6 g. We're gonna multiply that by the molar mass of oxygen gas which is going to be g, giving us 2.55 moles of oxygen gas. And then what we're going to do then is we're going to use that number to figure out how many moles of action dioxide that we can find from that. So we're going to take our To find our moles of our nitrogen dioxide. We're going to take our 2.55 moles of oxygen gas and multiplied by our multiple ratio. Once again, you can say that for every one mole of our oxygen gas, we have two moles of our nitrogen dioxide. And when our units cancel, we're left with 5.1 moles of nitrogen di oxide. And so we know that our limiting reactant is going to be our nitric oxide since it produces the least amount of our di nitrogen nitrogen dioxide. And so if we know our chemical reaction that we had followed by the entropy change here, we can calculate the change in entropy. So the change in our entropy Is essentially going to be equal to the moles that we found from our limiting reactant, which was our 2.4 moles of our nitrogen nitric oxide. We're gonna multiply by the conversion factor that we that for every minus or negative 1 16. kg joules. We have one mole And this is going to give us negative 135.9 killing jewels. Using this, we can then lastly calculate the change in our energy or internal energy to be specific. So are changing our internal energy is going to be equal to the change in entropy minus what is essentially our work. And so we already calculated what we had for work. So what we're gonna do is we're gonna combine our values, we're gonna take our negative 1 35.9 kg jewels. Subtract that by our minus 7.60 kg jewels. And this is going to give us a value of negative kg joules as our final answer. So these two are final answers for our work and our change in our internal energy. And with that we've answered the question overall. I hope this helped. And until next time.

Verified Solution

Key Concepts

Enthalpy Change (ΔH)

Work Done (PV Work)

Internal Energy Change (ΔE)