The standard enthalpy of the decomposition of N 2 O 4 to NO 2 is 58.04 kJ and standard entropy of this reaction is 176.7 J / K. Therefore the standard free energy change for this reaction at 25° C is

Question

The standard enthalpy of the decomposition of N 2 O 4 to NO 2 is 58.04 kJ and standard entropy of this reaction is 176.7 J / K. Therefore the standard free energy change for this reaction at 25° C is (A) -5.39 kJ (B) 539 kJ (C) 5.39 kJ (D) -539 kJ

Tardigrade Admin · Accepted Answer

Standard enthalpy of decompositions (Δ H) Standard entropy (Δ S)=176.7 J / K =176.7 × 10-3 kJ / K and temperature (T)=25° C =298 K Change in free energy (Δ G)=Δ H-T Δ S =58.04-298 ×(176.7 × 10-3) =58.04-52.65=5.39 kJ

Q. The standard enthalpy of the decomposition of $N_{2} O_{4}$ to $N O_{2}$ is $58.04 k J$ and standard entropy of this reaction is $176.7 J / K$ . Therefore the standard free energy change for this reaction at $2 5^{\circ} C$ is

$- 5.39 k J$

$539 k J$

$5.39 k J$

$- 539 k J$

Standard enthalpy of decompositions $(Δ H)$

Standard entropy $(Δ S) = 176.7 J / K = 176.7 \times 1 0^{- 3} k J / K$

and temperature $(T) = 2 5^{\circ} C = 298 K$

Change in free energy $(Δ G) = Δ H - T Δ S$

$= 58.04 - 298 \times (176.7 \times 1 0^{- 3})$

$= 58.04 - 52.65 = 5.39 k J$

Q. The standard enthalpy of the decomposition of N2​O4​ to NO2​ is 58.04kJ and standard entropy of this reaction is 176.7J/K. Therefore the standard free energy change for this reaction at 25∘C is

−5.39kJ

539kJ

5.39kJ

−539kJ