Question

For the decomposition of calcium carbonate, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): ΔH∘rxn 178.5kJ/mol ΔS∘rxn 161.0J/(mol⋅K) Calculate the temperature in kelvins above which this reaction is spontaneous. Express your answer to four significant figures and include the appropriate units.

Answer 1

Answer: $1.109\times 10^3$ with four significant digits.

Explanation:

Given,

$\Delta H$ = 178.5 KJ/mole = 178500 J/mole    (1kJ=1000J)

$\Delta S$ = 161.0 J/mole.K

According to Gibbs–Helmholtz equation:

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibbs free energy  

$\Delta H$ = enthalpy change  

$\Delta S$ = entropy change  

T = temperature in Kelvin

$\Delta G$= +ve, reaction is non spontaneous

$\Delta G$= -ve, reaction is spontaneous

$\Delta G$= 0, reaction is in equilibrium

As per question the reaction is spontaneous that means the value of  $\Delta G$ is negative or we can say that the value is less than zero.

Thus

$T\Delta S>\Delta H$

$T\times 161J/Kmol> 178500J/mol$

$T>1109K$

Significant figures are the figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Thus the temperature is $1.109\times 10^3$ in kelvins above which this reaction is spontaneous.