General Chemistry homework

Question

Consider the synthesis of ammonia in the gas phase:


N2(g) + 3 H2(g) → 2 NH3(g) ∆H°rxn = –91.8 kJ


Calculate ∆E (kJ) for this system if 125 g of ammonia is formed and the volume of the system decreases by 219 liters at a constant pressure of 1.50 atm.









































2. Consider the following reaction: 2 CH4(g) + O2(g) → 2 CH3OH(l). Use average bond enthalpies (bond energies) to estimate the heat of reaction, ∆Hrxn. See Table 9.3.





























3. Write the heat of formation chemical reaction for methanol CH3OH(l).
 





4. What mass of methanol, CH3OH(l), needs to combust with oxygen gas to heat 275 g of water from 20.0°C to 100.0°C? Assume there is only a 15.0% heating efficiency of the water from the combustion reaction and the combustion reaction produces H2O(l) not H2O(g). Hint: Write the balanced reaction then use Table 9.4 to find ∆Hrxn.

































































4.1. At constant pressure, what is the sign of the work (w) term for the methanol combustion reaction, +, –, or 0?

Answer 1

1.

To calculate ΔE for the system, first we need to calculate ΔH for this specific reaction. We start with 125 g NH3, and we know that there are about 17 g NH3 in 1 mol NH3. Furthermore, 91.8 kJ of energy are released for every 2 moles of NH3 put into the reaction. We can use stoichiometry to solve for ΔH, which is 125 g NH3 * (1 mol NH3 / 17 g NH3) * (-91.8 kJ / 2 mol NH3) = -337.5 kJ. Therefore, 337.5 kJ of energy were released by the reaction.

However, we are not done yet, as the air also did work on the system. The formula for pressure-volume work is W = -PΔV. P is constant at 1.50 atm. and ΔV is -219 L. as the volume of the system decreased by 219 liters. Therefore, W = -1.5 * -219 = 328.5 L*atm. There are approximately 0.101325 kJ in one L*atm, so this is equal to 33.2853 kJ.

Adding these two values together, we get -337.5 kJ + 33.2853 kJ = -304.215 kJ, but we only get three sig figs, so ΔE = -304 kJ.

2.

The method which we are using to find ΔHrxn in this case is practically breaking every single bond we had before and turning it into all the bonds we need in the end. This is helpful, as it takes about 413 kJ to break one mol of C-H bonds and creating one mol of C-H bonds releases about 413 kJ.

There are four C-H bonds in CH4, and there are 2 moles of CH4 in the equation. Therefore, to break apart every bond here, we need 8 * 413 kJ = 3304 kJ of energy. There is one double bond in O2 which takes 495 kJ of energy to break. On the other side, in CH3OH, there are three C-H bonds, one C-O bond, and one O-H bond, which release 413 kJ, 358 kJ, and 463 kJ respectively. Therefore, forming one mole of CH3OH releases 413 kJ * 3 + 358 kJ + 463 kJ = 2060 kJ of energy. However, the equation makes two moles of CH3OH, so we must multiply this number by 2 to reach 4120 kJ of energy.

As 4120 kJ of energy are released and 3799 of energy kJ are used, ΔHrxn = -4120 kJ + 3799 kJ = -321 kJ

3.

I wasn't completely sure what #3 was asking, so I think they are trying to say "What is ΔHrxn for methanol (l)?" This is basically checking our answer for the last question. To find this, we need to subtract the reactants' ΔH from the products' ΔH. ΔH for liquid methanol is -235.08 kJ/mol, and there are 2 moles of methanol in the equation. We need to subtract from this the ΔH for CH4 and O2, which are -74.533 kJ/mol and 0 kJ/mol respectively. There are 2 moles of CH4 in the equation as well.

ΔHrxn = 2 * -235.08 kJ - 2 * -74.533 kJ = -321.094 kJ. As we are only allowed three sig figs, ΔHrxn = -321 kJ

I'm sorry if this wasn't what #3 was asking for, but I didn't know what else it wanted.

4.

A very important thing about this reaction is that it's a combustion reaction. Combustion reactions always have something added to O2, and the products are always CO2 (g) and H2O. In this case, it is H2O (l). The unbalanced equation is CH3OH (l) + O2 (g) -> CO2 (g) + H2O (l). Balancing the equation, we find that 2 CH3OH (l) + 3 O2 (g) -> 2 CO2 (g) + 4 H2O (l).

To find ΔHrxn for this equation, we need to once again subtract the reactants from the products. The ΔH value for CH3OH(l), O2(g), CO2(g), and H2O(l) are -235.08 kJ/mol, 0 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol respectively. Putting it all together, we get ΔHrxn = 2 * -393.5 kJ + 4 * -285.8 kJ - 2 * -235.08 kJ = -1460.04 kJ However, the question tells us that only 15% of the reaction actually warms the water, so we're really only getting -1460.04 * 0.15 = -219 kJ per reaction.

Next, we need to find how much energy it takes to heat the water. We can use the equation q = m*csp*ΔT where m is the mass of water, csp is the specific heat capacity of water, and ΔT is the change in temperature. m = 275 g, csp = 4.184 J/gK, and ΔT = 80 degrees. Therefore, q = 275 * 4.184 * 80 = 92048 J = 92.048 kJ. Now we know that we need about 92 kJ of energy to heat the water.

Quick plug here, but for more thermochemistry fun, check out this YouTube video where I looked at some chemistry in the mushroom kingdom! I used the method I just showed you in this video, so that's there if you ever want more practice with thermo. https://youtu.be/QETSAvV_zuw

We can use stoichiometry again to convert from kJ of energy to g CH3OH. 2 moles of CH3OH produces 219 kJ, and there are 32 grams in one mol of CH3OH. 92.048 kJ * (2 mol CH3OH / 219 kJ) * (32 g CH3OH / 1 mol CH3OH) = 26.9 g CH3OH. We need 26.9 grams of methanol to combust with oxygen gas in order to heat the water.

4.1.

As discussed earlier, W = -PΔV. In this case, pressure is constant, so we only need to worry about the change in volume. Gas moles take up the most space, so looking back at the combustion reaction, we can see that there are three moles of gas in the reactants and two moles of gas in the products. We are decreasing the amount of gas, so we don't need as much volume to hold it. Since volume decreases, ΔV < 0. A negative times a negative is a positive, so W > 0, and the sign of work is +.

I hope this helped! Thanks for reading through it!