½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS
Answer:
B)−6,942 J
/mol
Explanation:
At constant temperature and pressure, you cand define the change in Gibbs free energy, ΔG, as:
ΔG = ΔH - TΔS
Where ΔH is enthalpy, T absolute temperature and ΔS change in entropy.
Replacing (25°C = 273 + 25 = 298K; 25.45kJ/mol = 25450J/mol):
ΔG = ΔH - TΔS
ΔG = 25450J/mol - 298K×108.7J/molK
ΔG = -6942.6J/mol
Right solution is:
<h3>B)−6,942 J
/mol</h3>
The coefficient for NaNO₃ = 6
<h3>Further explanation
</h3>
Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:
• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
Reaction
AI(NO₃)₃ +Na₂SO₄ →
Al₂(SO₄) +
NaNO₃
give coefficient
aAI(NO₃)₃ +bNa₂SO₄ →
Al₂(SO₄)₃ +c
NaNO₃
Al, left=a, right=2⇒a=2
N, left=3a, right=c⇒3a=c⇒3.2=c⇒c=6
Na, left=2b, right=c⇒2b=c⇒2b=6⇒b=3
The equation becomes :
2AI(NO₃)₃ +3Na₂SO₄ →
Al₂(SO₄)₃ +6NaNO₃
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