A) increase in amount of\[S{{O}_{3}}\]
B) increase in partial pressure of\[{{O}_{2}}\]
C) increase in the partial pressure of\[S{{O}_{2}}\]
D) change in equilibrium constant
E) none of the above
Correct Answer: A
Solution :
For the chemical reaction \[\underbrace{\underset{2\,mol}{\mathop{2S{{O}_{2}}}}\,+\underset{1\,mol}{\mathop{{{O}_{2}}}}\,}_{3\,mol}\xrightarrow{{}}\underset{2\,mol}{\mathop{2S{{O}_{3}}}}\,\] Here the volume is decrease therefore, increasing the total pressure leads to increase in the amount of\[S{{O}_{3}}\].You need to login to perform this action.
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