Question

# Consider the reaction equilibrium:$$2SO_2+O_2\rightleftharpoons 2SO_3$$ : $$\Delta{H}^0$$= -195 kJ on the basis of Le Chateliers principle, the condition favourable for the forward reaction is?

A
Increasing temperature as well as pressure
B
Lowering the temperature and increasing the pressure
C
Any value of temperature and pressure
D
Lowering of temperature as well as pressure

Solution

## The correct option is B Lowering the temperature and increasing the pressure$$\begin{array}{l} 2SO_{2}+O_{2} \rightleftharpoons \quad 2 \mathrm{SO}_{3} \quad \Delta \mathrm{H}^{\circ}=-195 \mathrm{~K} \mathrm{~J} \\ \text { It is exothermic reaction } \\ \text { so lowering temperature favours forward } \\ \text { reaction according to lechatliers principle } \\ \end{array}$$$$\begin{array}{l} \Delta n=2-(2+1)=-1 \\ \text { no. of moles decreases then increase in } \\ \text { Pressure favours forward reaction. } \\ \text { so (B) is correct } \end{array}$$Chemistry

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