The molar heat capacity of ethane is represented in the temperature range 298 k to 400 k by the empirical expression cp, m in j k1 mol 14.73 + (0.1272 t in k). the corresponding expressions for c(e) and h2(g) are given in the back of the atkins textbook. calculate the standard enthalpy of formation of ethane at 373 k from its value at 298 k, in kj mol

-88.66 kJ/mol

Explanation:

The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:

C(s):  Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)

H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)

Cp = A + BT + CT⁻²

For the Kirchoff's Law:

ΔHf = ΔH°f +

Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f  for ethene is -84.68 kJ/mol and the reaction is:

2C(s) + 3H₂(g) → C₂H₆

So, DCp:

dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83

dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788

dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵

dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²

= -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)

ΔHf = -84.68 - 3.80

ΔHf = -88.66 kJ/mol

answer is: nuclear fusion produces elements that are heavier than helium.

nuclei combine to form a heavier nucleus, releasing energy.  

for example, fusion of hydrogen atoms at the sun.  

this fusion nuclear reaction is in the inner core of the sun and hydrogen (atomic number 1) is conveted into helium (atomic number 2).  

the inner core is the central region of the sun and all solar energy is produced in the core by nuclear fusion.