ΔG deg will be negative above 7.27e+3 K.
<u>Explanation:</u>
- The ΔG deg with the temperature can be found using the formula and the formula is given below
- ΔG deg = ΔH deg - T ΔS deg
- Given data, ΔH deg = 181kJ and ΔSdeg=24.9J/K
- -T ΔS deg will be always negative and ΔG deg = ΔH deg will be positive and ΔG deg will be negative at relatively high temperatures and positive at relatively low temperatures
- solving the equation and substitute ΔGdeg=0
- ΔGdeg = ΔHdeg - T ΔSdeg
- T= ΔHdeg/ΔSdeg
- T=181 kJ / 2.49e-2 kJK-1
- By simplification we get
- T=7.27 × 10^3 K.
- Therefore, Go will be negative above 7.27 × 10^3 K
- Since ΔG deg = -RT lnK, when ΔGdeg < 0, K > 1 so the reaction will have K > 1 above 7.27 × 10^3 K.
- ΔG deg will be negative above 7.27e+3 K.
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1.) Air can be separated into its constituents such as oxygen, nitrogen etc.
2.) Air has a variable composition because at different places different amount of gases are present in air.
Answer:
<u>The new pressure is 1.0533 atm</u>
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Explanation:
According to<u> Boyle's Law :</u> The Pressure of fixed amount of gas is inversely proportional to Volume at constant temperature.
PV = Constant
P1V1 = P2V2
.....(1)
P1 = 3.16 atm
Accprding to question ,
V1 = V
V2 = 3 V
Insert the value of V1 , V2 and P1 in the equation(1)
V and V cancel each other
Hydrogen - 7.44%
Carbon - (100-7.44)% = 92.56%
Lets take 100 g of benzene, then we have
Hydrogen - 7.44 g
Carbon - 92.56 g
n - number of moles
n(H) = 7.44g *1 mol/1.0g = 7.44 mol
n(C) = 92.56 g* 1mol/12.0 g ≈ 7.713 mol
n(C) : n(H) = 7.713 mol : 7.44 mol = 1:1
Empirical formula is CH.
M(CH) = (12.0+ 1.00) g/mol = 13.0 g/mol
M (benzene) = 78.1 g/mol
M (benzene)/M(CH)= 78.1 g/mol/13.0 g/mol = 6
So, molecular formula of benzene is C6H6.
