mol-i and For C(graphite)-C(diamond) at 298 K and one bar, Δ,Ho-+1.895 kJ A,S® =-3.363 JK-l spontaneous?...
The standard Gibbs free energy for the transformation of diamond to graphite at 298 K is -2.9 kJ/mol. Why are diamonds not spontaneously transformed to graphite? C(diamond)→C(graphite) Δ?0=−2.9kJ/mol
Consider the reaction at 298 K. C(graphite)+2 H 2 (g)⟶ CH 4 (g)Δ?°=−74.6 kJ C(graphite)+2H2(g)⟶CH4(g)ΔH°=−74.6 kJ Calculate the quantities. Delta Ssys=_______ J/K Delata Ssurr =_______ J/K
What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K? Cdiamond?Cgraphite Express your answer to three significant figures and include the appropriate units. Gibbs free energy is a measure of the spontaneity of a chemical reaction. It is the chemical potential for a reaction, which is minimized at equilibrium. It is defined as G=H?TS Elemental carbon usually exists in one of two forms: graphite or diamond. It is generally believed that diamonds last...
What is the value of ?K for this aqueous reaction at 298 K? A+B↽−−⇀C+D Δ?°=26.52 kJ/mol
Consider the following reaction at 298 K. AH' - -74.6 kJ and AS C(graphite) + 2 H2(g) - CH (8) Calculate the following quantities. -80.8 J/K ASxys - + TOOLS AS univ- X10 Is this reaction spontaneous? Ово O yes
The standard enthalpy change for the following reaction is -111 kJ at 298 K. C(s,graphite) + 1/2 O2(g) --> CO(g) ΔH° = -111 kJ What is the standard enthalpy change for the reaction at 298 K? 2 CO(g) --> 2 C(s,graphite) + O2(g) Answer in kJ
Consider the reaction at 298 K. C(graphite) + 2Cl2(8) — CCI (1) AH° = -139 kJ Calculate the quantities. ASsys = J/K S TOOLS A TOOLS x102 ASsurr = asyr = O JK J/K ASuniy = Asuniv = UK J/K
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all speies. For the reactionC2H6(g)+H2(g)↽−−⇀2CH4(g)the standard change in Gibbs free energy is Δ𝐺∘=−32.8 kJ/mol.ΔG°=−32.8 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are 𝑃C2H6=0.400 bar,𝑃H2=0.150 bar, and 𝑃CH4=0.850 bar?𝑃CH4=0.850 bar? Δ𝐺=_____ kJ/molSk
What is the value of ? for this aqueous reaction at 298 K? A+B↽−−⇀C+DΔ?°=14.14 kJ/mol A + B ↽ − − ⇀ C + D Δ G ° = 14.14 kJ / mol ?= K =
What is the value of ? for this aqueous reaction at 298 K? A+B↽−−⇀C+DΔ?°=29.94 kJ/mol A + B ↽ − − ⇀ C + D Δ G ° = 29.94 kJ / mol