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Standard entropy change4/11/2023 ![]() ![]() The -ve sign of ΔS o indicates that the system becomes more ordered, which one might have expected given that two gases are becoming one gas in the reaction. It would have been impossible to be confident of correctly predicting the sign of ΔS for this reaction as there are 1 mole of a solid and 1 mole of a gas on both sides of the equation. The +ve sign of the entropy agrees with the observed direction of spontaneity. The entropy for oxygen in the gas mixture would be greater. Mixed and remain in the gas phase, they have less order as theyĪre more randomly distributed than before they were mixed. On the other hand, if two different gases are ![]() When a gas dissolves in water, its moleculesĪssociate with water molecules and this results in an increasedĭegree of order for the system - hence the entropy value is smaller Is smaller than that for gas phase water at the same temperature. Consequently the entropy value for liquid phase water The water molecules have much more freedom to move and are thus moreĭisordered. hence they have more entropy, so the sign of their entropies atĪs the temperature increases, the atoms of copper move more rapidlyĪnd hence are more disordered. Of the elements increase their motions and thus become more disordered The enthalpy of formation of oxygen gas is, byĭefinition, the enthalpy change for the processĪs the temperature increases, the component atoms and molecules More disordered due to their faster motion at the higher temperature. K going to 298 K, a process which involves its molecules becoming The entropy change is that for oxygen gas at 0 The reaction equation is H 2(g) + I 2(g) 2HI(g)Īnd another mixture of the same gases at 298 K is introduced, theĭeduce the direction of the reaction under these conditions.Īt 298 K, calculate the thermodynamic equilibrium constant.Ĭalculate the value of K for the reaction in Q 12 above.Īssuming standard conditions, deduce the temperature at which theĬalculate (i) D G o and (ii) K for the following reaction at standard conditions: In which direction will the reaction proceed? Δ G o for the conversion of graphite to diamondĪ mixture of hydrogen, iodine and hydrogen iodide gases is placed Of free energy of formation to calculate Δ G o Using entropy and enthalpy of formation tables,ĭeduce whether the following reaction is spontaneous under standardĪt 1 atmosphere, liquid and vapour phase water are in equilibriumĪt a temperature = 373 K represented by the equation Using standard entropy tables, calculate ΔS o at 298 K for the spontaneous reaction How would the entropy change for oxygen gas if When oxygen gas dissolves in water, its entropy is reduced. Mol -1 but for gaseous water at the same temperature, Inspection of the tables of standard entropy values of the elementsĮxplain the trend shown in the following values of S oįor liquid water at 298 K, S o = 69.9 J K -1 ![]() Shortcut to Questions Q: 1 2 3 4 6 7 8 9 10 11 ![]()
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