Question

$aA+bB\rightleftharpoons cC+dD$
In the above reaction, low pressure and high-temperature conditions shift the equilibrium in the backward direction, then which of the following is true?
(a)- (a + b) > (c + d), $\Delta H>0$
(b)- (a + b) < (c + d), $\Delta H>0$
(c)- (a + b) < (c + d), $\Delta H<0$
(d)- (a + b) > (c + d), $\Delta H<0$

Answer 1

Hint: The property of pressure can be used to calculate the number of moles of reactants and products. The property of temperature can be used to calculate the change in enthalpy ($\Delta H$) of the reaction.

Complete step by step answer:
In the reaction:
$aA+bB\rightleftharpoons cC+dD$

A, B, C, and D are the compounds that have a, b, c, and d number of moles respectively. The question says that low pressure and high-temperature conditions shift the equilibrium in the backward direction. The property of pressure can be used to calculate the number of moles of reactants and products. The property of temperature can be used to calculate the change in enthalpy ($\Delta H$) of the reaction.
So, low pressure will cause the reaction to occur in a backward reaction which means that high pressure will shift the reaction towards forward or right. Therefore, according to Le Chatelier’s principle, high pressure favors when the number of moles decreases. This means that (a + b) > (c + d).
Now, the high temperature causes the reaction to occur in a backward reaction which means that low temperature will shift the reaction towards forward or right. Low temperature is only accompanied when the reaction is exothermic. The reaction is exothermic means that the change in enthalpy is negative, so $\Delta H<0$.
So, the correct answer is “Option D”.

Note: There is another factor that can decide the nature of the reaction i.e., concentration. The reaction will shift forward if the concentration of the reactant is increased and the reaction will shift backward if the concentration of the product increases.