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It's a calorimetry calculation. Here's how you do it. EXAMPLE When 25.0 mL of 0.700 mol/L NaOH was mixed in a calorimeter with 25.0 mL of 0.700 mol/L HCl, both initially at 20.0 °C, the temperature increased to 22.1 °C. The heat capacity of the calorimeter is 279 J/°C. What is the molar enthalpy of neutralization per mole of HCl? Solution The equation for the reaction is NaOH + HCl → NaCl ...
How would you calculate the enthalpy change, delta H, for the process in which 33.3 g of water is converted from liquid at 4.6 C to vapor at 25.0 C? For water: H = 44.0 kJ/mol at 25.0 C and s = 4.18J/g C for H2O(l).
The explanation below offers an approach to the enthalpy of fusion given the specific heat of the substance in question in both its solid state and in the liquid state among other information. This approach to the enthalpy change of fusion requires the following information: Mass of the sample; Specific heat of the substance under solid and liquid state c_{"solid"} and c_{"liquid"}; The ...
Many enthalpy changes are difficult to measure directly under standard conditions, enthalpy of formation being such a case. It is a lot easier to measure the enthalpy of combustion using calorimetry. Since the elements and the compound from which they are made will have the same products of combustion we can set up an energy cycle.
Since enthalpy is a state function, we only need to know the beginning and end states of the reaction, and we can figure out the standard enthalpy for the reaction. Standard enthalpies of formation DeltaH_f^@ are tabulated in Thermodynamics tables you should be able to find in your textbook appendix. With those, we can construct the following equation basically looking at the enthalpy required ...
In a coffee cup calorimeter, 50.0 mL of 1.0 M NaOH and 50.0 mL of 1.0 M HCl are mixed. Both solutions are originally at 23.5 degrees C and after the reaction, the final temperature is 29.9 degrees Celsius. Calculate the enthalpy change for the neutralization of HCl by NaOH
You use the standard enthalpy of the reaction and the enthalpies of formation of everything else. For most chemistry problems involving ΔH_f^o, you need the following equation: ΔH_(reaction)^o = ΣΔH_f^o(p) - ΣΔH_f^o(r), where p = products and r = reactants. EXAMPLE: The ΔH_(reaction)^o for the oxidation of ammonia 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) is -905.2 kJ. Calculate ΔH_f ...
DeltaH = H_f - H_i Values of H (enthalpy) for particular reactants or reactions will always be given in the exercise. However, depending on the unit, you may be forced to either multiply H by moles (when unit is Kj/mol) or leave it as it is (when unit is Kj). At first, you count H for finals, then for "ingredients" (by addition) and substract results. I think it is the easiest way. EDIT: That ...
Hess's law states that the total enthalpy change does not rely on the path taken from beginning to end. So, you can calculate the enthalpy as the sum of several small steps. There are a few rules that you must follow when manipulating an equation. You can reverse the equation. This will change the sign of #ΔH#.
How can enthalpy change be determined for an aqueous solution? How does enthalpy change with pressure? How do you calculate standard molar enthalpy of formation?