Chemistry, 29.03.2020 21:48, mulan8382

How can an unknown A Heaction be determined using Hess's law?
A. The free energy of the reaction is used to determine the A H for
the reaction.
O
B. The reaction is repeated at different temperatures to determine
the A Hreaction
O
C. Enthalpies from reaction steps are added to determine an
unknown A Hreaction:
D. The unknown A Heaction is determined after the reaction is run in a
calorimeter.

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answered: uhhgray

Choice C. In Hess's Law, $\Delta H_\text{reaction}$ from reaction steps are combined to determine an unknown

Explanation:

Consider this example: the $\Delta H_\text{reaction}$ for the two reactions are given:

Reaction (1): $\rm A + B \to C + D$ . $\Delta H_\text{reaction} = \Delta H(1)$ .Reaction (2): $\rm C + D \to E + F$ . $\Delta H_\text{reaction} = \Delta H(2)$ .

The goal is to find the $\Delta H_\text{reaction}$ for reaction (3):

$\rm A + B \to E + F$ .

Note that adding reaction (1) and (2) gives:

$\rm A + B + C + D \to C+ D + E + F$ .

Eliminate the intermediary species $\rm C$ and $\rm D$ to get:

$\rm A + B \to E + F$ .

That's the same as reaction (3). With the help of Hess's Law, it is possible to find the $\Delta H_\text{reaction}$ for reaction (3) without taking any measurement.

Since Reaction (1) + Reaction (2) = Reaction (3), by Hess's Law,

$\begin{aligned}& \Delta H_\text{reaction}(3)\\ &= \Delta H_\text{reaction}(1) + \Delta H_\text{reaction}(2)\end{aligned}$ .

No Gibbs Free energy $\Delta G$ is involved in this calculation. There's not even the need to carry out an experiment or take any new measurements. Because of that, Hess's Law can be very useful for finding the $\Delta H_\text{reaction}$ of reactions that are hard to carry out, control, or measure. That includes finding the lattice enthalpy of ionic compounds, and finding the $\Delta H_\text{reaction}$ of the hard-to-control reaction $\displaystyle \rm C\, (\text{s, graphite}) + \frac{1}{2} \,O_2\, (g) \to CO\, (g)$ .

answered: Guest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.