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deltaH=28 kJ/mol, deltaS=0.109 kJ(molK)
Wiki User
∙ 7y ago
Sierra Murtha
Wiki User
∙ 7y agoFor the rx to be spontaneous, ∆Gº has to be negative and ∆Gº = ∆Hº - T∆Sº so substitute in the values of ∆H and solve for ∆S or plug in ∆S and solve for ∆H, or plug in both and solve for T. The question as stated is unlcear as to what is given and what needs to be calculated.
What form a spontaneous reaction at 298 K?
H = 28 kJ/mol, S = 0.109 kJ/(molK)
Which statement describes a reaction at 298 K if H 31 kJ mol S 0.093 kJ molK?
It is not spontaneous.
Which direction of the reaction equilibrium is favored at 298 K room temperature use the reaction I2 solid to I2 gas when H 62.4 kJmol and S 0.145 kJmolK?
I2(s) --> I2(g); dH=62.4kJ/mol; dS=0.145kJ/mol. The reaction will favor the product at this temperature. Your entropy is positive and your enthalpy is also positive, so this reaction will not be spontaneous at all temperatures. But at room temperature, which is 298K, it will be spontaneous and proceed left to right. (this is the sublimation of iodine)
Use the reaction I2s I2 g H 62.4 kJ mol S 0.145 kJ molK for Question 10. Which direction of the reaction is favored at 298 K room temperature?
Toward I2(s) production
Which direction of the reaction is favored at 298 K?
Toward I2(s) production
What form a spontaneous reaction at 298 K?
H = 28 kJ/mol, S = 0.109 kJ/(molK)
What statement describes a reaction at 298 K if H 31 kJmol S 0.093 kJ(molK)?
It is not spontaneous.
Which statement describes a reaction at 298 K if H 31 kJ mol S 0.093 kJ molK?
It is not spontaneous.
Which direction of the reaction equilibrium is favored at 298 K room temperature use the reaction I2 solid to I2 gas when H 62.4 kJmol and S 0.145 kJmolK?
I2(s) --> I2(g); dH=62.4kJ/mol; dS=0.145kJ/mol. The reaction will favor the product at this temperature. Your entropy is positive and your enthalpy is also positive, so this reaction will not be spontaneous at all temperatures. But at room temperature, which is 298K, it will be spontaneous and proceed left to right. (this is the sublimation of iodine)
Use the reaction I2s I2 g H 62.4 kJ mol S 0.145 kJ molK for Question 10. Which direction of the reaction is favored at 298 K room temperature?
Toward I2(s) production
What phase is it at room temperature if an element has a melting point above 298 kelvin?
it would be in ice form
Which direction of the reaction is favored at 298 K?
Toward I2(s) production
Which compound is formed from its elements by an exothermic reaction at 298 K and 101.3 kPa?
H2O(g)
What is 298 over 1000 in simplest form?
149/500
Which equation represents an exothermic reaction at 298 K?
C(s)+O2(g) -> CO2(g)
What are some non- examples of endothermic process?
You are given the following chemical rection N 2 ( g ) O 2 ( g ) → 2 N O ( g ) Provide an explanaition for why this reaction is endothermic (both conceptual, and mathematical); Is this reaction spontaneous at 298 K? If not, at what temperature does it become spontaneous? Data given: Δ H ∘ f = 90.4 kJ/mol for N O and Δ S reaction = 24.7 J/K Let's start with the math to get it out of the way. A reaction is said to be endothermic if its change in enthalpy, Δ H reaction , is positive. We can calculate this change in enthalpy from what the data provides us. Δ H reaction = 2 moles NO ⋅ 90.4 k J m o l − ( 1 mole N 2 ) ⋅ 0 k J m o l − ( 1 mole O 2 ) ⋅ 0 k J m o l Δ H reaction = 2 moles NO ⋅ 90.4 k J m o l = 180.8 k J The trick here was to be aware of the fact that the enthalpy of formation ( Δ H ∘ f ) for elements is zero. Since Δ H reaction > 0 , the reaction is indeed endothermic. Conceptually, this reaction is endothermic despite the fact that a bond (between N and O ) is formed; this happens because the N 2 molecule has its two atoms bonded together by a very strong triple bond, which means that more energy must be put into breaking this bond than is released when the N O molecule is formed. Now, in order for a reaction to be spontaneous, the sign of Δ G the Gibbs free energy - must be negative at the given temperature. We can therefore determine this reaction's spontaneity by using Δ G reaction = Δ H reaction − T ⋅ Δ S reaction Δ G reaction = 180.8 ⋅ 10 3 J − 298 K ⋅ 24.71 ⋅ J K = 173.4 k J The reaction is not spontaneous at this temperature. We can determine at what temperature the reaction starts to be spontaneous by setting Δ G reaction = 0 . 0 = Δ H reaction − T ⋅ Δ S reaction T = Δ H reaction Δ S reaction = 180.8 ⋅ 10 3 J 24.71 J K = 7317 K This reaction becomes spontaneous at 7317 K . As a conclusion, questions about endothermic or exothermic processes revolve around Δ H , Δ S , and Δ G if a reaction's spontaneity is in question.
How much is 20 percent less of 298 rupees?
20 percent less of 298 rupees would be 238.40 rupees.
