Question

What is the increase in entropy when 11.2 L of O₂ is mixed with 11.2 L of H₂ at STP?

a) 0.576 J/K
b) 5.76 J/K
c) 7.56 J/K
d) 2.76 J/K

Answer 1

The increase in entropy when mixing 11.2 L of O₂ with 11.2 L of H₂ at STP is 5.76 J/K, corresponding to answer (b).

Step-by-step explanation:

The increase in entropy when 11.2 L of O₂ is mixed with 11.2 L of H₂ at STP can be calculated using the formula for the entropy change of mixing ideal gases: ΔS = -nR(sum of mole fraction times the natural log of the mole fraction of each component). At standard temperature and pressure (STP), one mole of an ideal gas occupies 22.4 L.

Therefore, 11.2 L of each gas corresponds to 0.5 moles. Since both gases have the same amount, the mole fraction for each is 0.5. Plugging these values and the gas constant R (8.314 J/mol·K) into the formula gives:

ΔS = -nR[(0.5 * ln(0.5)) + (0.5 * ln(0.5))]

ΔS = -0.5 mol * 8.314 J/mol·K * [2 * (-0.693)]

ΔS = 5.76 J/K

Therefore, the correct answer to the question of the increase in entropy when mixing 11.2 L of O₂ with 11.2 L of H₂ at STP is 5.76 J/K. This option corresponds to answer (b).