The reaction 2SO_2 + O_2 rightleftharpoons 2S | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$; $\Delta H = -ve$. Since $\Delta H$ is negative,the reaction is exothermic.

Vedclass · Accepted Answer

Low temperature,high pressure

Vedclass · Answer

(B) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$; $\Delta H = -ve$. Since $\Delta H$ is negative,the reaction is exothermic. According to Le Chatelier's principle,for an exothermic reaction,a low temperature favours the forward reaction. Regarding pressure,the number of moles of gaseous reactants is $2 + 1 = 3$,and the number of moles of gaseous products is $2$. Since the number of moles decreases in the forward direction,high pressure favours the forward reaction. Therefore,the reaction is favoured by low temperature and high pressure.

The reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$; $\Delta H = -ve$ is favoured by

Similar Questions

For the formation of $NH_{3(g)}$ from its constituent elements,the favourable conditions for its formation are

The effect of the addition of helium gas to the following reaction in an equilibrium state is:
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$

In the equilibrium $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} + 22 \ kcal$,the formation of ammonia is favored by:

Consider the equilibrium ${N_2}_{(g)} + 3{H_2}_{(g)} \rightleftharpoons 2N{H_3}_{(g)}$,$\Delta H = -93.6 \ kJ$. The maximum yield of ammonia is obtained by:

For the reaction: $A + B + Q \rightleftharpoons C + D$,if the temperature is increased,then the concentration of the products will

Vedclass Test Series

Exam Paper Generator

Online Exam Module