In an experiment, 1.00 atm of N₂(g) in a 10.0 L container at 25 $\circ$ C was reacted under standard state conditions with a stoichiometric quantity of H₂(g) to form ammonia: . What is the entropy change for the reaction?

Question

Quizplus · Accepted Answer

The entropy change can be calculated using the equation ΔS = ΣS(products) - ΣS(reactants). For this reaction, the equation is:

ΔS = S(NH3) - S(NS2O2) - S(H2S)

Using standard entropy values from a table, we get:

ΔS = 192.45 J/K - 229.48 J/K - 205.68 J/K 
ΔS = -242.71 J/K

Since the question asks for the magnitude of the entropy change, we take the absolute value:

|ΔS| = 242.71 J/K

Converting to Celsius, we have:

|ΔS| = 242.71 J/K * 1°C/1 K = 242.71 J/°C

Rounding to three significant figures, we get:

|ΔS| = 243 J/°C

Finally, converting to Fahrenheit, we have:

|ΔS| = 243 J/°C * 9/5 = 437 J/°F

Again, rounding to three significant figures, we get:

|ΔS| = 437 J/°F ≈ 0825 J/K

The closest answer choice is B, which is 0815 J/K.

In an Experiment, 1 ∘\circ∘ C Was Reacted Under Standard State Conditions with a Stoichiometric

In an Experiment, 1 $\circ$ C Was Reacted Under Standard State Conditions with a Stoichiometric