Find the temperature at which K_p = 42.0 for the reaction H₂(g)+ I₂(g) $\rarr$ 2HI(g). [Given: at 25°C, for H₂(g), $\Delta$H°_f = 0, S° = 131.0 J/mol·K; for I₂(g), $\Delta$H°_f = 62.26 kJ/mol, S° = 260.6 J/mol·K; for HI(g), $\Delta$H°_f = 25.9 kJ/mol, S° = 206.3 J/mol·K; assume that $\Delta$H° and $\Delta$S° are independent of temperature.]

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The correct answer is A.The expression for the equilibrium constant Kp is:Kp = (PH2I)^2 / (PH2 * PI2)The change in standard Gibbs free energy (ΔG°) is related to the equilibrium constant Kp by the equation:ΔG° = -RT ln KpThe change in standard Gibbs free energy can also be expressed in terms of the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°) by the equation:ΔG° = ΔH° - TΔS°Combining these equations, we get:-RT ln Kp = ΔH° - TΔS°Solving for T, we get:T = ΔH° / (ΔS° + R ln Kp)Substituting the given values, we get:T = (62.26 kJ/mol + 25.9 kJ/mol) / (206.3 J/mol·K + 260.6 J/mol·K + 8.314 J/mol·K ln 42.0)T = 1040 K

Find the Temperature at Which Kp = 42 →\rarr→ 2HI(g) [Given: at 25°C, for H2(g)

Find the Temperature at Which K_p = 42 $\rarr$ 2HI(g) [Given: at 25°C, for H₂(g)