Q. 194.2( 33 Votes )

# For the reaction

2 A(g) + B(g) → 2D(g)

= –10.5 kJ and = –44.1 JK^{–1}.

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

Answer :

∆U = -10.5 kJ

Also, ∆U = ∆H – ∆nRT

Where,

∆U = change in energy

∆H = change in enthalpy

∆n = moles of products – moles of reactants = 2-3 = -1

R = 8.314 × 10^{-3} kJ mol^{-1} K^{-1} - constant

T = temperature; here room temperature is taken as 298 K

Substituting all the above values, we get

∆H = -10.5kJ + (-1 mol × (8.314× 10^{-3}kJ mol^{-1} K^{-1}) × 298 K

= -12.977kJ

we know that

∆G = ∆H - T∆S, where

∆G = change in Gibbs energy

∆H = change in enthalpy, got as -12.977 kJ

∆S = change in entropy, given as -44.1 JK^{-1} = -44.1 × 10^{-3} kJK^{-1}

T = temperature at which reaction occurs = 298K

Substituting the above values

∆G = -12.977 – (298 × -44.1 × 10^{-3})

= -12.977 – ( -13.14)

= 0.163kJ

As ∆G > 0, reaction doesn’t occur spontaneously. Reactions are spontaneous only when ∆G < 0.

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