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What would be the final value for the enthalpy of reaction you use for this intermediate reaction C2H4 plus 3 O2 2 CO2 plus 2 H2O H -1410 kJ?
The final value for the enthalpy of reaction for the combustion of ethylene (C2H4) in your given reaction is -1410 kJ. This indicates that the reaction is exothermic, releasing 1410 kJ of energy as products (2 CO2 and 2 H2O) are formed from the reactants (C2H4 and 3 O2). Therefore, the enthalpy change, ΔH, for the complete combustion of ethylene is -1410 kJ.
At which temperature would a reaction with h -92 kj mol s-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. For a reaction to be spontaneous, ΔG must be less than 0. Given ΔH = -92 kJ/mol and ΔS = -199 kJ/(mol·K), we can set up the equation -92 kJ/mol - T(-199 kJ/(mol·K)) < 0. Solving for T gives T > 0.462 K, indicating that the reaction will be spontaneous at temperatures above this value.
Ask us anythingIf you need to multiply the following reaction by 2 to be an intermediate reaction in a Hess's law problem what would be the final value for the enthalpy of reaction you use for this in?
When you multiply a reaction by a factor, you also multiply the enthalpy change (ΔH) of that reaction by the same factor. Therefore, if you multiply the reaction by 2, you would take the original enthalpy of reaction and multiply it by 2. For example, if the original ΔH is -100 kJ, the final value for the enthalpy of reaction you would use would be -200 kJ.
In a chemical reaction the reactants contain 385 kJ of chemical energy and the products contain 366 kJ of chemical energy?
In this chemical reaction, the reactants have 385 kJ of chemical energy, while the products have only 366 kJ. This indicates that the reaction is exothermic, as it releases energy in the form of heat, resulting in a net loss of 19 kJ of energy. The difference represents the energy released to the surroundings during the reaction.
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.
If you need to reverse the following reaction and multiply it by 2 in order for it to be an intermediate reaction in a Hess's law problem what would be the final value for the enthalpy of reaction yo?
2820 kJ
What would be the final value for the enthalpy of reaction you use for this intermediate reaction?
Can you please provide me with the specific reaction or context for which you need the enthalpy value?
If you need to reverse the following reaction in order for it to be an intermediate reaction in a Hess's law problem what would be the final value for the enthalpy of rea?
286 kJ
What would be the final value for the enthalpy of reaction you use for this intermediate reaction C2H4 plus 3 O2 2 CO2 plus 2 H2O H -1410 kJ?
The final value for the enthalpy of reaction for the combustion of ethylene (C2H4) in your given reaction is -1410 kJ. This indicates that the reaction is exothermic, releasing 1410 kJ of energy as products (2 CO2 and 2 H2O) are formed from the reactants (C2H4 and 3 O2). Therefore, the enthalpy change, ΔH, for the complete combustion of ethylene is -1410 kJ.
What is the activation energy of peroxodisulphate and iodide reaction?
From an experiment I did in my chemistry lab, I got a value of 52.0 kJ. But I have no idea what the "true" Ea is.
What begins the chemical reaction of a safety match?
ikjhhmhj,kj,kj
At which temperature would a reaction with h -92 kj mol s-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. For a reaction to be spontaneous, ΔG must be less than 0. Given ΔH = -92 kJ/mol and ΔS = -199 kJ/(mol·K), we can set up the equation -92 kJ/mol - T(-199 kJ/(mol·K)) < 0. Solving for T gives T > 0.462 K, indicating that the reaction will be spontaneous at temperatures above this value.
Ask us anythingIf you need to multiply the following reaction by 2 to be an intermediate reaction in a Hess's law problem what would be the final value for the enthalpy of reaction you use for this in?
When you multiply a reaction by a factor, you also multiply the enthalpy change (ΔH) of that reaction by the same factor. Therefore, if you multiply the reaction by 2, you would take the original enthalpy of reaction and multiply it by 2. For example, if the original ΔH is -100 kJ, the final value for the enthalpy of reaction you would use would be -200 kJ.
What is the heat of reaction of methane in steam reforming?
The heat of reaction for methane steam reforming is an endothermic process, meaning heat is required for the reaction to occur. The reaction produces hydrogen and carbon monoxide from methane and steam. The heat of reaction for this process is around 206 kJ/mol of methane converted.
In a chemical reaction the reactants contain 385 kJ of chemical energy and the products contain 366 kJ of chemical energy?
In this chemical reaction, the reactants have 385 kJ of chemical energy, while the products have only 366 kJ. This indicates that the reaction is exothermic, as it releases energy in the form of heat, resulting in a net loss of 19 kJ of energy. The difference represents the energy released to the surroundings during the reaction.
If you need to reverse the following reaction in order for it to be an intermediate reaction in a Hess's law problem what would be the final value for the enthalpy of reaction you use for this in?
The final value for the enthalpy of the reverse reaction used in a Hess's law problem would simply be the negative of the original value of the enthalpy of the forward reaction. This is because reversing a reaction changes the sign of the enthalpy change.
If you need to reverse the following reaction and multiply it by 2 in order for it to be an intermediate reaction in a Hess's law problem what would be the final value for the enthalpy of reaction you?
If you need to reverse a reaction and multiply it by 2 in Hess's law, the enthalpy change of the reaction will also change sign and double in magnitude. This is because reversing a reaction changes the sign of the enthalpy change. Multiplying the reaction by a factor also multiplies the enthalpy change by that factor. Therefore, the final value for the enthalpy of the reaction will be twice the original magnitude but with the opposite sign.
