From the D E given for 25 ° C, calculate D H at the same temperature for the reaction:
2 HI (g) →H₂ (g) + I₂ (g) D E = +9.48 kJ

Question

From the D E given for 25 ° C, calculate D H at the same temperature for the reaction: 2 HI (g)→H₂ (g) + I₂ (g) D E = +9.48 kJ

Quizplus · Accepted Answer

The enthalpy change, DH, is related to the internal energy change, DE, by the equation DH = DE + RTΔn, where R is the gas constant, T is the temperature in Kelvin, and Δn is the change in moles of gas between reactants and products. In this case, Δn = (2 moles of gas in products) - (1 mole of gas in reactants) = 1, since we count the stoichiometric coefficients as the number of moles.

Plugging in the given values, we get DH = +9.48 kJ + (8.314 J/(mol·K))(298 K)(1) / 1000 J/kJ = +9.80 kJ.

However, we need to round our final answer to two significant figures, since that's the precision of the given DE value. The closest choice is B, +9.48 kJ, which is the same as the given DE value. Therefore, B is the best choice.

From the D E Given for 25 ° C, Calculate