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A reaction will be spontaneous at 298 K if the Gibbs free energy change (ΔG) for the reaction is negative. This means that the reaction will proceed in the forward direction without requiring an external input of energy. The equation ΔG = ΔH - TΔS can be used to determine if a reaction is spontaneous at a given temperature, where ΔH is the change in enthalpy and ΔS is the change in entropy.
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What statement describes a reaction at 298 K if H 31 kJmol S 0.093 kJ(molK)?
It is not spontaneous.
What state is oxygen at 298 K?
Oxygen is a gas at 298 K.
What temperature would a reaction with H -92 kJmol S -0.199 kJ(molK) be spontaneous?
To determine the temperature at which the reaction becomes spontaneous, we can use the Gibbs free energy equation: ΔG = ΔH - TΔS. A reaction is spontaneous when ΔG < 0. Given ΔH = -92 kJ/mol and ΔS = -0.199 kJ/(mol·K), we set ΔG to 0 and solve for T: 0 = -92 kJ/mol - T(-0.199 kJ/(mol·K)). This simplifies to T = 462.31 K. Thus, the reaction is spontaneous at temperatures above approximately 462 K.
Which direction of the reaction is favored at 298 K (room temperature)?
Toward I2(s) production
At which temperature would a reaction with h-92 kjmol s-0.199 kj(molk) be spontaneous?
To determine whether the reaction is spontaneous, we can use the Gibbs free energy equation, ( \Delta G = \Delta H - T\Delta S ). For the reaction to be spontaneous, ( \Delta G ) must be less than 0. Given ( \Delta H = -92 , \text{kJ/mol} ) and ( \Delta S = -0.199 , \text{kJ/(mol K)} ), we can set up the inequality ( -92 , \text{kJ/mol} - T(-0.199 , \text{kJ/(mol K)}) < 0 ). Solving this will give the temperature threshold above which the reaction becomes spontaneous.
What form a spontaneous reaction at 298 K?
A spontaneous reaction at 298 K is one in which the Gibbs free energy change (ΔG) is negative. This means that the reaction is energetically favorable and will proceed in the forward direction without the need for external energy input.
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.
What would form a spontaneous reaction at 298 K?
deltaH=28 kJ/mol, deltaS=0.109 kJ(molK)
Which compound is formed from its elements by an exothermic reaction at 298 K and 101.3 kPa?
Carbon dioxide (CO2) is formed from its elements (carbon and oxygen) by an exothermic reaction at 298 K and 101.3 kPa.
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)
Which equation represents an exothermic reaction at 298 K?
An exothermic reaction is one where heat is released to the surroundings. An example of an exothermic reaction equation at 298 K is: 2H2(g) + O2(g) -> 2H2O(l) + heat
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?
At 298 K, the direction of a reaction is favored based on whether it is exothermic or endothermic. If the reaction is exothermic, it is favored in the direction that consumes heat, while for an endothermic reaction, it is favored in the direction that produces heat. The reaction will proceed in the direction that helps to minimize the overall energy of the system.
Using the Gibbs free energy equation what is the temperature range for which the oxidation of lead is spontaneous?
The reaction is spontaneous below 554.8/0.1975 K.
What state is oxygen at 298 K?
Oxygen is a gas at 298 K.
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.
