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 .
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The final volume of the balloon is 84.75 ml
<u><em>calculation</em></u>
The volume is calculated using Charles law formula
that is V1/T1= V2/T2 where,
V1= 5.00 x10^2 ml
T1= 22.0 °c into kelvin = 22 + 273 = 295 K
V2= ? ml
T2 =0.500 x 10^2 K
make V2 the subject of the formula by multiplying both side by T2
V2 is therefore = T2V1/T1
V2 = (0.500 x 10^2 K x 5.00 x10^2 ml) / 295 K = 84.75 ml
The molar mass of a 0.314-gram sample of gas having a volume of 1.6 l at 287 k and 0.92 atm. is 5.23 gram.
<h3>What is an Ideal Gas ?</h3>
An Ideal Gas is a law for ideal gas which states that the product of Pressure and Volume is equal to the product of the moles of sample, universal gas constant and temperature.
PV = nRT
The given data in the question is
Pressure = 0.92 atm
Volume = 1.6 l
Temperature = 287 K
0.92 * 1.6 = n * 287 * 0.082
n = 0.06 moles
0.06 = 0.314 / molecular weight
molecular weight = 0.314/0.06
Molecular weight = 5.23 gm
Molar mass = 5.23 gm.
The temperature and pressure conditions at which the gas behaves like an ideal gas is 273 K and 1 atm.
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Answer:
The genetic variation in the population will decrease or stay the same
Explanation: