Explanation:
Equation of the reaction:
Br2(l) + Cl2(g) --> 2BrCl(g)
The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.
The standard enthalpy change of formation for a compound,
ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.
This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction
1/2Br2(g) + 1/2Cl2(g) → BrCl(g)
Here, ΔH°rxn = ΔH°f
This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl
Using Hess' law,
ΔH°f = total energy of reactant - total energy of product
= (1/2 * (+112) + 1/2 * (+121)) - 14.7
= 101.8 kJ/mol
ΔH°rxn = 101.8 kJ/mol.
Answer:
10 kg of ice will require more energy than the released when 1 kg of water is frozen because the heat of phase transition increases as the mass increases.
Explanation:
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In this case, since the melting phase transition occurs when the solid goes to liquid and the freezing one when the liquid goes to solid, we can infer that melting is a process which requires energy to separate the molecules and freezing is a process that releases energy to gather the molecules.
Moreover, since the required energy to melt 1 g of ice is 334 J and the released energy when 1 g of water is frozen to ice is the same 334 J, if we want to melt 10 kg of ice, a higher amount of energy well be required in comparison to the released energy when 1 kg of water freezes, which is about 334000 J for the melting of those 10 kg of ice and only 334 J for the freezing of that 1 kg of water.
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Answer:
hypothesis
Explanation:
a question that can be answered wity a hypothesis is a question a scientist can answer
Answer:
Part A is just T2 = 58.3 K
Part B ∆U = 10967.6 x C
You can work out C
Part C
Part D
Part E
Part F
Explanation:
P = n (RT/V)
V = (nR/P) T
P1V1 = P2V2
P1/T1 = P2/T2
V1/T1 = V2/T2
P = Pressure(atm)
n = Moles
T = Temperature(K)
V = Volume(L)
R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.
bar = 0.986923 atm
N = 14g/mol
N2 Molar Mass 28g
n = 3.5 mol N2
T1 = 350K
P1 = 1.5 bar = 1.4803845 atm
P2 = 0.25 bar = 0.24673075 atm
Heat Capacity at Constant Volume
Q = nCVΔT
Polyatomic gas: CV = 3R
P = n (RT/V)
0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))
V = (nR/P) T
V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K
V = (0.28721/1.4803845) x 350
V = 0.194 x 350
V = 67.9036 L
So V1 = 67.9036 L
P1V1 = P2V2
1.4803845 atm x 67.9036 L = 0.24673075 x V2
100.52343693 = 0.24673075 x V2
V2 = P1V1/P2
V2 = 100.52343693/0.24673075
V2 = 407.4216 L
P1/T1 = P2/T2
1.4803845 atm / 350 K = 0.24673075 atm / T2
0.00422967 = 0.24673075 /T2
T2 = 0.24673075/0.00422967
T2 = 58.3 K
∆U= nC ∆T
Polyatomic gas: C = 3R
∆U= nC ∆T
∆U= 28g x C x (350K - 58.3K)
∆U = 28C x 291.7
∆U = 10967.6 x C
