Answer:
The reaction releases energy
Explanation:
The products of an exergonic reaction have a lower energy state (Delta-G) compared to the reactants. Therefore there is a negative delta –G between products and reactants after the reactions. This means some energy is lost into the environment usually through light or heat.
Exergonic reactions are characterized by a net release of energy but they still require a small initial energy input to start, referred to as the 'activation energy'. The speed or direction of the reaction is not determined by whether it's exergonic.
In the context of chemical reactions, the true statement for all exergonic reactions is that such reactions result in a net release of energy. However, even exergonic reactions, which are characterized by energy release, require a small initial input of energy to get started. This initial energy demand is referred to as the 'activation energy'. Also, it's important to note that the speed of the reaction or its directionality (whether it proceeds only in a forward direction) are not inherently determined by whether a reaction is exergonic. These aspects depend on other reaction conditions and catalysis.
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argon (Ar)
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Argon (Ar) has a complete valence electron shell.
The element that has a complete valence electron shell is argon (Ar).
Answer:
1.7 × 10⁴ J
Explanation:
Step 1: Calculate the heat required to raise the temperature of ice from -15 °C to 0°C
We will use the following expression.
Q₁ = c(ice) × m × ΔT
Q₁ = 2.03 J/g.°C × 25 g × [0°C - (-15°C)] = 7.6 × 10² J
Step 2: Calculate the heat required to melt 25 g of ice
We will use the following expression.
Q₂ = C(fusion) × m
Q₂ = 80. cal/g × 25 g × 4.184 J/1 cal = 8.4 × 10³ J
Step 3: Calculate the heat required to raise the temperature of water from 0°C to 75 °C
We will use the following expression.
Q₃ = c(water) × m × ΔT
Q₃ = 4.184 J/g.°C × 25 g × (75°C - 0°C) = 7.8 × 10³ J
Step 4: Calculate the total heat required
Q = Q₁ + Q₂ + Q₃
Q = 7.6 × 10² J + 8.4 × 10³ J + 7.8 × 10³ J = 1.7 × 10⁴ J
Answer:
d. Ar, because of its higher effective nuclear charge
For the secon part see explanation below.
Explanation:
The first ionization energy is the energy required to remove an electron from the atom from its outermost shell. It depends on the nuclear charge, distance from the nucleus and the screening of other electrons in the inner shells of the atom.
Comparing Cl and Ar we see that being both elements of the third period, the Ar atom has one more proton than Cl and therefore the electron feels more nuclear charge making the first ionization of Ar greater than Cl.
a) False, electronegativity relates to attraction for an electron and not to the first ionization.
b) False, again electron affinity is not first ionization, it is defined as the energy released when the atom captures an added electron.
c) False,athough it is true that Ar has a complete octet, the higher first ionization is affected by nuclear charge. The screening of electrons in the n= 1 and 2 shells is almost the same so what is important is that the electrons in the n= 3 shell feel more nuclear charge.
d) True for all the reasons given previously : the higher effective nuclear charge in Ar.
For the second part, we have to make an inventory of the bonds being broken and formed:
ΔHºrxn = H broken - H formed, where H is the bond energy
H2 C = CH_2 + H-Br ⇒ CH_3CH_2Br
ΔHºrxn = ( 1 C=C + 4 C-H + 1 H-Br) - ( 1 C-C + 5 C-H + 1 C-Br)
ΔHºrxn (kJ) = (614 + 4(413) + 363) - ( 347 + 5 (413) + 276)
ΔHºrxn (kJ) = 2629 - 2688 = -59 kJ
This value is not in the choices due to mistaken bond energy values from the tables.
Answer:
1. Ar, because of its higher effective nuclear charge.
2. ∆Hrxn = -200 KJ/mol
Explanation:
The size of the atoms of chemical elements can be measured from their atomic radius which is also affected by the effective nuclear charge.
Recall that elements in a particular period have the same number of electron shells. Also, along a given period, atomic radius decreases due to an increase in the effective (positive) nuclear charge. This is because as the atomic (proton) number increases along that period, the charge on the nucleus also increases. With more protons in the nucleus the overall attraction between the positively charged nucleus and the negatively charged surrounding electrons increases, so the electrons are pulled closer to the nucleus thereby leading to a decrease in the atomic size.
So, along a given period atomic size decreases due to an increase in the effective nuclear charge.
The first ionization energy is the minimum energy (in kilojoules) needed to strip one mole of electrons from one mole of a gaseous atom of an element to form one mole of a gaseous unipositive ion.
Along a particular period, ionization energy increases due to an increase in the effective nuclear charge and a decrease in atomic radius. This is because, the smaller the atom the more stable it is and the more difficult it will be to remove an electron.
For the second question,
The enthalpy change of a reaction is the difference in the bond dissociation energies of the reactants and products. Bonds are broken in reactant molecules and formed in product molecules. Bond breaking energies are usually intrinsic ( endothermic, +be ∆H ) while bond forming energies are usually extrinsic ( exothermic, -ve ∆H ).
So,
∆Hrxn = n∆H(reactants/bonds broken) - m∆H(products/bonds formed)
Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation.
First, draw the correct Lewis structures of the compounds.
Next, identify all the bonds broken and formed.
Then, from the bond dissociation energies ( usually given or looked up in texts ), sum up the bond breaking energies and the bond forming energies and subtract the bond forming energies from the bond breaking energies.
Considering this equation:
H_2C = CH_2 + H-Br rightarrow CH_3CH_2Br
The equation is balanced.
Bonds broken (number of bonds ):
I. C=C (1)
II. H-Br (1)
III. C-H (4)
Bonds formed:
I. C-C (1)
II. C-H (5)
III. C-Br (1)
∆Hrxn = [ ( 1 x C=C ) + ( 4 x C-H ) + ( 1 x H-Br ) ] – [ ( 1 x C-C ) + ( 5 x C-H ) + ( 1 x C-Br ) ]
∆Hrxn = [ ( 1 x 614 ) + ( 4 x 413 ) + ( 1 x 141 ) ] – [ ( 1 x 348 ) + ( 5x 413 ) + ( 1 x 194 ) ]
∆Hrxn = [ ( 614+1652+141) ] – [ ( 348 + 2065 + 194 ) ]
∆Hrxn = 2407 – 2607
∆Hrxn = -200KJ/mol
Answer: A typical hydrogen fuel cell produces 0.5 V to 0.8 V per cell. To increase the voltage individual cells can be connected in series.
Answer: 40.496%
Hope this helps! (: