Solution:
For the given response,
\[\begin{array}{*{35}{l}}
2\text{ }A\left( g \right)\text{ }+\text{ }B\left( g \right)\text{ }\to \text{ }2D\left( g \right) \\
~ \\
ng\text{ }=\text{ }2\text{ }\text{ }\left( 3 \right)\text{ }=\text{ }\text{ }1\text{ }mole \\
\end{array}\]
Subbing the worth of ∆Uθ in the declaration of ∆H:
\[\begin{array}{*{35}{l}}
H\theta \text{ }=\text{ }U\theta \text{ }+\text{ }ngRT \\
~ \\
=\text{ }\left( \text{ }10.5\text{ }kJ \right)\text{ }\text{ }\left( \text{ }1 \right)\text{ }\left( 8.314\text{ }\times \text{ }103\text{ }kJ\text{ }K1\text{ }mol1\text{ } \right)\text{ }\left( 298\text{ }K \right) \\
~ \\
=\text{ }\text{ }10.5\text{ }kJ\text{ }\text{ }2.48\text{ }kJ\text{ }H\theta \text{ }=\text{ }\text{ }12.98\text{ }kJ \\
\end{array}\]
Subbing the upsides of ∆Hθ and ∆Sθ in the declaration of ∆Gθ :
\[\begin{array}{*{35}{l}}
G\theta \text{ }=\text{ }H\theta \text{ }\text{ }TS\theta \\
~ \\
=\text{ }\text{ }12.98\text{ }kJ\text{ }\text{ }\left( 298\text{ }K \right)\text{ }\left( \text{ }44.1\text{ }J\text{ }K1\text{ } \right) \\
~ \\
=\text{ }\text{ }12.98\text{ }kJ\text{ }+\text{ }13.14\text{ }kJ \\
~ \\
G\theta \text{ }=\text{ }+\text{ }0.16\text{ }kJ \\
\end{array}\]
Since ∆Gθ for the response is positive, the response won’t happen unexpectedly.