Question

Question 6.19 For the reaction

2A(g) + B(g)→2D(g)

ΔUθ = –10.5 kJ and ΔSθ= –44.1 JK–1.

Calculate ΔGθ for the reaction, and predict whether the reaction may occur spontaneously.

Class XI Thermodynamics Page 183

Answer 1

                    2A(g) + B(g) -------> 2D(g) : ΔU° = -10.5 KJ
Δng = np - nr 
       = 2 - (2 + 1) = 2 - 3 = -1 

ΔH° = ΔU° + ΔΔngRT,  from formula,
       = -10.5 KJ + (-1) × 0.008314 KJ/K/mol × 298K
       =-10.5 KJ - 2.447 KJ
       = -12.947 KJ
now, use the formula, ΔG° = ΔH° - TΔS°
                                           = -12.947 KJ - 298K × (-44 × 10⁻³ J/K)
                                           = -12.947 KJ + 13.14 KJ
                                           = +0.165 KJ
the reaction will not occur spontanteously because ΔG° is positive.
      

Answer 2

Hi

Here is your answer,

2A(g) + B(g) → 2D ; Δng = np - nr = 2-3 = -1 

  ΔH° = ΔU° + ΔngRT

  ΔH° = -10.5 kJ + ( -1 × 8.314 × 10⁻³ kJ mol⁻¹ × 298K )

  ΔH° = -10.5 + (- 2.477) kJ mol⁻¹

  ΔH°= -12.977 kJ mol⁻¹¹

  ΔG° = ΔH° - TΔS°

  ΔG° = -12.977 kJ mol⁻¹ - (298 K × -44.1 × 10⁻³ kJ⁻¹ mol⁻¹)

  ΔG° = -12.977 kJ mol⁻¹ + 13.14 kJ mol⁻¹ = + 0.165 kJ mol⁻¹



∴ The reaction will not occur spontaneously because ΔG° is positive.


Hope it helps you !