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