5.73 Consider the reaction N2(g) + 3H2(g) ⟶ 2NH3(g) Δ H =...

5.73 Consider the reaction $N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)$ $\Delta H = -92.6 \text{ kJ/mol}$ When 2 mol of $N_2$ react with 6 mol of $H_2$ to form 4 mol of $NH_3$ at 1 atm and a certain temperature, there is a decrease in volume equal to 98 L. Calculate $\Delta U$ for this reaction. (The conversion factor is 1 L \cdot atm = 101.3 J.)

Transcript

00:01 All right, so for this question, we're given the following chemical reaction, and we're told it has a change in enthalpy enthalpy associated with it of negative 92 .6 kilojoules per mole.

00:09 And i like to say kilojoules per mole of reaction, meaning that this is the change in enthalpy associated with one mole of n2 gas plus three moles of h2 gas, yields two moles of nh3 gas.

00:23 So we're told that we're using this reaction, but we actually generate four moles of it.

00:29 So we actually generate four moles of nh3 gas.

00:38 And we know that per two moles of nh3 gas, we have a change in enthalpy of negative 92 .6 kilojoules.

00:48 So the kilojoules for this specific reaction, or the change in enthalpy of this specific reaction, is going to equal negative 185 .2 kilojoules.

01:01 Now we also know in a constant pressure experiment, which this is, the pressure is at 1 atm, the change in enthalpy is also equal to the heat.

01:10 In this case, this is an exothermic process and 185 .2 kilojoules of heat regenerated.

01:17 Now, this is helpful because we're trying to find the change in internal energy of this particular reaction, 4 moles of nh3 gas are generated.

01:28 And we know that the change in internal energy is equal to the heat plus the work done.

01:33 Now, we have our heat, but we don't have our work.

01:39 We know that work is equal to the negative pressure times the change in volume.

01:43 And we know both of these.

01:45 You know that the pressure is one atmosphere, and that the volume decreased by 98 liters...

Question

Please give Ace some feedback