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What is the difference between the heat (q) of a reaction and the reaction enthalpy (Hrxn)?
The heat (q) of a reaction is the amount of energy transferred as heat during a chemical reaction, while the reaction enthalpy (Hrxn) is the overall change in heat energy of a reaction at constant pressure. The main difference is that heat (q) is the actual energy transferred, while reaction enthalpy (Hrxn) is a measure of the total heat change in a reaction.
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What is hrxn for chemical reaction co2g 2kohs h2og k2co3s?
The balanced chemical equation is: CO2(g) + 2KOH(s) -> H2O(g) + K2CO3(s). The enthalpy change (ΔHrxn) for this reaction is the enthalpy of formation of the products minus the enthalpy of formation of the reactants. To calculate it, you would need the standard enthalpies of formation for all species involved.
Determine the heat of reaction for these chemical reactions. Are they endo- or exothermic 1. C s H 2 O g CO g H 2. I cant seem to get the same answer twice. Can anyone help me?
C(s) + H2O(g) --> CO(g) + H2(g) I assume this is the reaction you have written. In order to find if the reaction is endo- or exothermic we can use Hess's law to estimate the Heat of reaction. Hess's Law uses the standard heat of formation of the products and reactants in the reaction to determine the heat of reaction. If the heat of reaction is negative then the reaction is exothermic, if it is posititve then the reaction is endothermic. Hrxn = SUM(coefficients*Hf(products)) - SUM(coefficients*Hf(reactants)) The heat of formation of all elements is zero using Hess's law, so all we need is the Hf of water(g) & carbon monoxide (g), which is -241.83 & -110.52 kJ/mol respectively. (You can find these Hf values in most Chemistry of Chemical Engineering Handbooks.) Since the above reaction is stoichiometrically correct we can calculate the Hrxn as follows: Hrxn = {(1)*(0) + (1)*(-110.52 kJ/mol)} - {(1)*(-241.83 kJ/mol)+(1)*(0)} = +131.31 kJ/mol Since the Hrxn is positive the above reaction is endothermic.
Use average bond energies to calculate delta Hrxn for the combustion of ethanol?
The combustion of ethanol is 2C2H5OH(l) + 6O2(g) -> 4CO2(g) + 6H2O(l). The average bond energies are: C-C = 348 kJ/mol, C-O = 358 kJ/mol, C=O = 805 kJ/mol, O-H = 463 kJ/mol. Calculate the energy required to break the bonds in the reactants and form the bonds in the products. Then subtract the energy required to break the bonds from the energy released in forming the new bonds to find the delta Hrxn.
What temperature is required to produce a carbon dioxide partial pressure of 1.0 ATM?
The temperature required to produce a carbon dioxide partial pressure of 1.0 ATM depends on the volume of the container and the total pressure in the system. Using the ideal gas law equation, you can calculate the temperature needed.
The combustion of 0.4196 g of a hydrocarbon containing only C and H releases 17.55 kj of heat the masses of the products are CO2 equals 1.219 g and H2O equals 0.290 g?
whats is the empirical formula of the compound? if the approximate molar mass of the compound is 76g/mol, calculate its standard enthalpy of formation ANSWER 1.419 g CO_2_ * 1 mol CO_2_ /44.01 g CO_2_ *1mol C/1 mol CO_2_ = 0.03224 mol C 0.290 H_2_O * 1mol H_2_O/18.02 g H_2_O * 2 mol H/1mol H_2_O=0.03219 MOL H Ratio is 1:1 so empirical formula is CH=13g/mol molar mass 76g/mol ->76/13=6 molecular formula C_6_H_6_ 2C_6_H_6_+15O_2_----->12CO_2_ + 6H_2_O 17.55kj/0.4196 C_6_H_6_* 78.11G C_6_H_6_/1mol C_6_H_6_ * 2 mol C_6_H_6_= 653 kj/mol (releases) Hrxn=12H_f_CO_2_ + 6H_f_ H_2_O - 2 H_f_ C_6_H_6_ - 15 -6534kj/mol=(120(-39305kj/mol)+6(-285.8kj/mol)-(2)(H_f_ C_^_H_^_)-(15) (0) H_f_(C_6_H_6_)=49kj/mol
What is the enthalpy of a reaction?
The amount of energy that is used or released as heat in a reaction.
What is hrxn for chemical reaction co2g 2kohs h2og k2co3s?
The balanced chemical equation is: CO2(g) + 2KOH(s) -> H2O(g) + K2CO3(s). The enthalpy change (ΔHrxn) for this reaction is the enthalpy of formation of the products minus the enthalpy of formation of the reactants. To calculate it, you would need the standard enthalpies of formation for all species involved.
Determine the heat of reaction for these chemical reactions. Are they endo- or exothermic 1. C s H 2 O g CO g H 2. I cant seem to get the same answer twice. Can anyone help me?
C(s) + H2O(g) --> CO(g) + H2(g) I assume this is the reaction you have written. In order to find if the reaction is endo- or exothermic we can use Hess's law to estimate the Heat of reaction. Hess's Law uses the standard heat of formation of the products and reactants in the reaction to determine the heat of reaction. If the heat of reaction is negative then the reaction is exothermic, if it is posititve then the reaction is endothermic. Hrxn = SUM(coefficients*Hf(products)) - SUM(coefficients*Hf(reactants)) The heat of formation of all elements is zero using Hess's law, so all we need is the Hf of water(g) & carbon monoxide (g), which is -241.83 & -110.52 kJ/mol respectively. (You can find these Hf values in most Chemistry of Chemical Engineering Handbooks.) Since the above reaction is stoichiometrically correct we can calculate the Hrxn as follows: Hrxn = {(1)*(0) + (1)*(-110.52 kJ/mol)} - {(1)*(-241.83 kJ/mol)+(1)*(0)} = +131.31 kJ/mol Since the Hrxn is positive the above reaction is endothermic.
How do you determine the change in temperature and final temperature of an endothermic reaction?
The amount of heat generated or absorbed in a chemical reaction can be studied using a calorimeter. our chemical reaction) is placed in a well-insulated vessel surrounded by water (surroundings). A thermometer is used to measure the heat transferred to or from the system to the surroundings. The heat that the chemical reaction puts out, or takes up, (qrxn) is simply the moles of the limiting reagent, nlimiting reagent times ΔHrxn. qrxn = nlimiting reagent·ΔH
Use average bond energies to calculate delta Hrxn for the combustion of ethanol?
The combustion of ethanol is 2C2H5OH(l) + 6O2(g) -> 4CO2(g) + 6H2O(l). The average bond energies are: C-C = 348 kJ/mol, C-O = 358 kJ/mol, C=O = 805 kJ/mol, O-H = 463 kJ/mol. Calculate the energy required to break the bonds in the reactants and form the bonds in the products. Then subtract the energy required to break the bonds from the energy released in forming the new bonds to find the delta Hrxn.
Why Natural Gas and Steam Reforming reaction is endothermic?
I assume this is the reaction you are talking about:CH4 + H2O (g) --> CO(g) + 3H2(g)From thermodynamics you can approximate the standard heat of reaction Hrxn with Hess' Law. This is the sum of the heats of formations of the products minus the sum of the heats of formations of the reactants in their stoichemtric ratios.Hrxn = SUM [(3)*Hf (H2)+(1)*Hf(CO)] - SUM [(1)*Hf (CH4) + (1)*Hf(H2O (g))]Looking up these values in a Chemical Engineering Handbook or textbook we can substitue and find the heat of reaction.Hrxn = SUM [(3)*(0 kJ/mol) + (1)*(-110.52 kJ/mol)] - SUM [(1)*(-74.85 kJ/mol) +(1)*(-241.83 kJ/mol)]= -110.52 - (-316.68) kJ/mol= + 206.16 kJ/mol, therefore the reaction is endothermicRemember standard heats of formation of elements, as hydrogen in this case, are zero.
What temperature is required to produce a carbon dioxide partial pressure of 1.0 ATM?
The temperature required to produce a carbon dioxide partial pressure of 1.0 ATM depends on the volume of the container and the total pressure in the system. Using the ideal gas law equation, you can calculate the temperature needed.
The combustion of 0.4196 g of a hydrocarbon containing only C and H releases 17.55 kj of heat the masses of the products are CO2 equals 1.219 g and H2O equals 0.290 g?
whats is the empirical formula of the compound? if the approximate molar mass of the compound is 76g/mol, calculate its standard enthalpy of formation ANSWER 1.419 g CO_2_ * 1 mol CO_2_ /44.01 g CO_2_ *1mol C/1 mol CO_2_ = 0.03224 mol C 0.290 H_2_O * 1mol H_2_O/18.02 g H_2_O * 2 mol H/1mol H_2_O=0.03219 MOL H Ratio is 1:1 so empirical formula is CH=13g/mol molar mass 76g/mol ->76/13=6 molecular formula C_6_H_6_ 2C_6_H_6_+15O_2_----->12CO_2_ + 6H_2_O 17.55kj/0.4196 C_6_H_6_* 78.11G C_6_H_6_/1mol C_6_H_6_ * 2 mol C_6_H_6_= 653 kj/mol (releases) Hrxn=12H_f_CO_2_ + 6H_f_ H_2_O - 2 H_f_ C_6_H_6_ - 15 -6534kj/mol=(120(-39305kj/mol)+6(-285.8kj/mol)-(2)(H_f_ C_^_H_^_)-(15) (0) H_f_(C_6_H_6_)=49kj/mol
