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Answer:
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C
Explanation:
<u>Step 1: </u>Data given
ΔH = −320.1 kJ/mol
ΔS = −86.00 J/K · mol.
<u>Step 2:</u> Calculate the temperature
ΔG<0 = spontaneous
ΔG= ΔH - TΔS
ΔH - TΔS <0
-320100 - T*(-86) <0
-320100 +86T < 0
-320100 < -86T
320100/86 > T
3722.1 > T
The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)
We can prove this with Temperature T = 3730 K
-320100 -3730*(-86) <0
-320100 + 320780 = 680 this is greater than 0 so it's non spontaneous
T = 3700 K
-320100 -3700*(-86) <0
-320100 + 318200 = -1900 this is lower than 0 so it's spontaneous
The temperature is quite high because of the big difference between ΔH and ΔS.
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C
Answer is: enthalpy for this reaction is -175.91 kJ.
Chemical reaction: NH₃ + HCl → NH₄Cl.
ΔrH = ∑ΔfH(products of reaction) - ∑ΔfH(reactants).
ΔrH = ΔfH(NH₄Cl) - (ΔfH(NH₃) + ΔfH(HCl)).
ΔrH = -314.4 kJ/mol · 1 mol - (-46.19 kJ/mol · 1 mol + (-92.30 kJ/mol) · 1 mol).
ΔrH = -314.4 kJ + 138.49 kJ.
ΔrH = -175.91 kJ.
3, 9, 16 are correct! 17 should be b because buffered means that it resists change in pH
Answer:
c
Explanation:
the correct answer would be answer c