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At which temperature would a reaction with AH = -92 kJ/mol, AS = -0.199kJ/(mol·K) be spontaneous?
the answer is 400
What else can I help you with?
What statement describes a reaction at 298 K if H 31 kJmol S 0.093 kJ(molK)?
It is not spontaneous.
What temperature would a reaction with H -220 kJmol and S -0.05 kJ(molK) be spontaneous?
3600 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.
At which temperature would a reaction with H -220 kJmol and S -0.05 kJ(mol K) be spontaneous?
To determine the temperature at which the reaction is spontaneous, we use the Gibbs free energy equation: ΔG = ΔH - TΔS. A reaction is spontaneous when ΔG < 0. Given ΔH = -220 kJ/mol and ΔS = -0.05 kJ/(mol K), we set up the equation: -220 kJ/mol - T(-0.05 kJ/(mol K)) < 0. Solving for T gives T > 4400 K, meaning the reaction is spontaneous at temperatures above 4400 K.
At which temperature would a reaction with h-220 kjmol and s-0.05?
To determine the temperature at which the reaction becomes spontaneous, we can use the Gibbs free energy equation: ΔG = ΔH - TΔS. For the reaction to be spontaneous, ΔG must be less than zero. Given ΔH = -220 kJ/mol and ΔS = -0.05 kJ/(mol·K), we can set up the equation: 0 = -220 kJ/mol - T(-0.05 kJ/(mol·K)). Solving for T gives T = 220 kJ/mol / 0.05 kJ/(mol·K) = 4400 K. Therefore, the reaction becomes spontaneous below this temperature.
At which temperature would a reaction with H -92 kJmol S -0.199 kJ(molK) be spontaneous?
The condition for a reaction to be spontaneous is ΔG < 0, where ΔG = ΔH - TΔS. At the temperature where ΔG becomes negative, the reaction will be spontaneous. You can calculate this temperature using the given values of ΔH and ΔS.
What can be said about a reaction with triangle H 620 kJmol and triangle S -046 kJ(mol.K)?
it is never spontaneous
What statement describes a reaction at 298 K if H 31 kJmol S 0.093 kJ(molK)?
It is not spontaneous.
What temperature would a reaction with H -220 kJmol and S -0.05 kJ(molK) be spontaneous?
3600 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.
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.
Why At which temperature would a reaction with H -220 kJmol and S -0.05 kJ(mol) be spontaneous?
To determine the temperature at which a reaction with an enthalpy change (ΔH) of -220 kJ/mol and an entropy change (ΔS) of -0.05 kJ/(mol·K) becomes spontaneous, we use the Gibbs free energy equation: ΔG = ΔH - TΔS. A reaction is spontaneous when ΔG < 0. Setting ΔG to zero for spontaneity and solving for T gives us T = ΔH/ΔS. Plugging in the values, T = -220 kJ/mol / -0.05 kJ/(mol·K) = 4400 K. Thus, the reaction is spontaneous at temperatures below 4400 K.
What can be said about a reaction with H 620 kJmol and S -0.46 kJ molK?
it can never be spontanious
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)
What can be said about a reaction with H -890 kJmol and S -0.24 kJ(mol K)?
The reaction is exothermic because the enthalpy change is negative (-890 kJ/mol). The reaction may be spontaneous at low temperatures due to the negative entropy change (-0.24 kJ/(mol K)), which decreases the overall spontaneity of the reaction.
What can be said about a reaction with delta H 62.4 kJmol and delta S 0.145 kJ(mol K)?
A reaction with a positive ΔH of 62.4 kJ/mol indicates that it is endothermic, absorbing heat from the surroundings. The positive ΔS of 0.145 kJ/(mol K) suggests that the reaction leads to an increase in entropy, favoring disorder. To determine the spontaneity at a given temperature, the Gibbs free energy change (ΔG = ΔH - TΔS) can be evaluated; if ΔG is negative, the reaction is spontaneous. At higher temperatures, the positive entropy change may make the reaction more favorable.
What can be said about a reaction with H 620 kJmol and S -0.46 kJ(molK)?
it can never be spontanious
