Calculate the temperature (in kelvins) at which the sign of AG° changes from positive to negative for the reaction below. This corresponds to the temperature at which K < 1 changes to K > 1. Assume that AH° and AS° are constant and do not change with temperature. CH4(g) + 202(g)→CO2(g) + AH° = -802 kJ/mol and AS° = -5.20 J/K 2H20(g) • mol T = K

Calculate the AH°, AS°, and AG° at 298 K for the following reaction: (Use the values found in Thermodynamic Properties to calculate your answers for questions a and b.) Al2O3(s) + 3 C(graphite) + 3 Cl₂(g) → 2 AICI 3(s) + 3 CO(g) 4.0 -64 a) ΔΗΟ kJ/mol b) AS° ✓ J/K.mol kJ/mol (Determine from the temperature, AH and AS) c) AGO 4.0 76.9 4.0 -89 (d) Is the reaction extensive at standard conditions and 298 K? O Yes O NO (e) Write the expression for K. (Format example: Kp = pHC12 / pH₂. pCl₂ would be entered as K_{p} = pHCI^{2}/pH_{2}. pCl_{2}.) Help chemPad XX (PCO)3 (pCl₂)³ ← Greek (PCO)^3/(pCl_2)^3 Your answer provides a different type of equation than was expected. (f) Calculate the value of K at 298 K 2.51e-16 X (Please answer to 3 significant figures.) (g) Estimate the value of K at 600 K. 1.79e-8 X (Please answer to 3 significant figures.)

What is AG for the reaction l2(g) → 2 1(g) at 25.0 °C if K = 6.1 x 10-2²? (R = 8.314 J/mol.K)

Ammonia can be produced by the reaction of hydrogen gas and nitrogen gas, as shown below: N2(g) + 3H2(g) → 2NH3(g) Given that the standard free energy of formation of NH3 (g) is -104 kJ/mol at 298 K, calculate the equilibrium constant, K, at this temperature. To express an answer in exponential notation, use E to indicate the exponent. For example, 3.0 x 103 would be written, 3.0E3.

For the reaction: CuS (s) + H₂(g) → H₂S (g) + Cu (s) Given: ΔS ⁰ (CuS) = 66.5 J/mole-K ΔHf ⁰ (CuS) = -53.1 kJ/mol ΔS ⁰ (H₂S) = 205 J/ mole-K ΔHf ⁰ (H₂S) = -20.6 kJ/mol ΔS ⁰ (Cu) = 33.3 J/ mole-K ΔH f ⁰ (Cu) = 0.0 kJ/mol ΔS ⁰ (H₂) = 131 J/ mole-K ΔH f ⁰(H₂) = 0.0 kJ/mol The standard free energy of the reaction ΔG ⁰ rxn is A) -37.9 kJ/mole B) 44.6 kJ/mole C) -20.3 kJ/mole D) +20.3kJ/mole and spontaneous E) +37.9 kJ/mole

Calculate the temperature at which the reaction below will be at equilibrium, then explain how this temperature can affect the spontaneity, meaning how using higher or lower than this temp can affect the value of delta G.   Given.      Br2(l) → Br2 (g),  ΔH = 31.0 kj/mol ΔS = 92.9 j/mol

Consider the reaction in the Haber process 3H2 (g) + N2 (g) → 2NH3 (g)where delta H° = -91.8 kJ/mol and delta S° = -197 J/(mol*K). What is the value of the equilibrium constant (K) at a temperature of 125°C?

alculate ΔS per mole for the melting of H2O at  0.°C,  ΔHfus = 6.0 kJ/mol.   J/mol·K

What is true about the following reaction at 25°C? H₂(g) + 2BrF(g) = 2HBr(g) + F₂(g) \Delta Hr\ deg = 115 kJ mol-¹ \Delta Sr\deg = 11.55 J mol-¹ K-¹ \Delta Gr\deg = 111 kJ mol-¹ i. The reaction is endothermic. ii. As the reaction proceeds heat is transferred from the surroundings to the system.