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Answer:
Keqq = 310
Note: Some parts of the question were missing. The missing values are used in the explanation below.
Explanation:
<em>Given values: ΔH° = -178.8 kJ/mol = -178800 J/mol; T = 25°C = 298.15 K; ΔS° = -552 J/mol.K; R = 8.3145 J/mol.K</em>
Using the formula ΔG° = -RT㏑Keq
㏑Keq = ΔG°/(-RT)
where ΔG° = ΔH° - TΔS°
㏑Keq = ΔH° - TΔS°/(-RT)
㏑Keq = {-178800 - (-552 * 298.15)} / -(8.3145 * 298.15)
㏑Keq = -14221.2/-2478.968175
㏑Keq = 5.73674166
Keq = e⁵°⁷³⁶⁷⁴¹⁶⁶
Keq = 310.05
Answer:
D. 2666.67g/dm3
Explanation:
Hello,
In this case, we compute the solubility as:
Thus, we compute:
Therefore, answer is D. 2666.67g/dm3.
Best regards.
The value of ΔG° (gibbs free energy change) is -2703kj and -2657kj.
Given ,
combustion reaction of butane :
C4H10(g) + 13/2O2 (g) →4CO2(g) + 5H2O(g)
Method-1 :
We know ,
ΔGrxn = sum of ΔG (product ) - sum of ΔG (reactant )
= [4 × (-394.4) + 5×(-228.6) ] - [1×(-16.7)]
ΔGrxn = -2720.6kj
Method-2 :
We know ,
ΔG = ΔH -TΔS
T =298K
Thus , ΔHrxn = sum of ΔH (product ) - sum of ΔH (reactant )
= [4×(-393.5) + 5×(-241.8) ] - [ 1×(-126)]
ΔHrxn = -2657kj
ΔSrxn = sum of ΔS (product ) - sum of ΔS( reactant )
= [ 4×(213.7) + 5×188.7 ] - [ 13/2 ×205 + 1×310 ]
ΔSrxn = 155.8j/K
ΔSrxn = 0.1558kj/K
Thus , ΔG = ΔH - TΔS
= -2657 - ( 298 × 0.1558kj/K )
ΔG = -2720kj
Hence , the value of ΔG is -2703kj .
Learn more combustion reaction here :
brainly.com/question/13251946
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