High temperature and high pressure (as per Le Chatelier's principle) favor,

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$N_{2} + 3 H_{2} ⇌ 2 N H_{3}, ∆ H = Q 1$

$3 O_{2} ⇌ 2 O_{3}, ∆ H = + Q 3$

$C a C O_{3} ⇌ C a O + C O_{2}, ∆ H = + Q 2$

$N_{2} + O_{2} ⇌ 2 N O, ∆ H = + Q 4$

answer is C.

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Detailed Solution

The reaction, $3 O_{2} ⇌ 2 O_{3}, ∆ H = + Q 3$

In the above reaction , see how there are 3 molecules in the reaction on the left side of the equation, but only 2 on the right. Le Chatelier's Principle states that the system will react by favoring the reaction that generates the fewest molecules. The pressure and temperature will increase once again as a result of that. As a result, the response will favor the goods more.

Therefore, the correct answer is option (C) $3 O_{2} ⇌ 2 O_{3}, ∆ H = + Q 3$