Question

For the reaction, \[\ce{X2O_{4(l)} -> 2XO_{2(g)}}\] ΔU = 2.1 kcal, ΔS = 20 cal K^–1 at 300 K Hence: ΔG is:

पर्याय

• 2.7 kcal

• – 2.7 kcal

• 9.3 kcal

• – 9.3 kcal

Answer 1

– 2.7 kcal

Explanation:

The change in Gibbs energy is given by ΔG = ΔH – TΔS

where; ΔH = change in enthalpy or reaction

ΔS = change in entropy of reaction

As a result, in order to calculate ΔG, the values of ΔH must be known. The equation can be used to compute the value of ΔH.

ΔH = ΔU + Δn_gRT

where; ΔU = change in internal energy

Δn_g = (no. of moles of gaseous products)-(no. of moles of gaseous reactants)

= 2 – 0 = 2

But; ΔH = ΔU + Δng

ΔU = 2.1 kCal = 2.1 × 10³ Cal  .....[∵ 1 kCal = 10³ Cal]

ΔH = (2.1 × 10³) + (2 × 2 × 300) = 3300 Cal

Hence; ΔG = ΔH – TΔS

ΔG = (3300) – (300 × 20) = – 2700

ΔG = – 2.7 kCal