Chemical And Ionic Equilibrium
- Chemical And Ionic Equilibrium Result Question 1
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- Chemical and Ionic Equilibrium - Result Question 1
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- Chemical and Ionic Equilibrium - Result Question 4
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- Chemical and Ionic Equilibrium - Result Question 7
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- Chemical and Ionic Equilibrium - Result Question 8
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- Chemical And Ionic Equilibrium Result Question 43 — JEE PYQ Chapterwise
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- Chemical and Ionic Equilibrium - Result Question 9
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- Chemical And Ionic Equilibrium Result Question 48 — JEE PYQ Chapterwise
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- Chemical And Ionic Equilibrium Result Question 43 — JEE PYQ Chapterwise
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- Chemical and Ionic Equilibrium - Result Question 10
- Chemical And Ionic Equilibrium Result Question 51 — JEE PYQ Chapterwise
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- Chemical and Ionic Equilibrium - Result Question 11
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- Chemical and Ionic Equilibrium - Result Question 50
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- Chemical and Ionic Equilibrium - Result Question 55
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 63-1
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
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Chemical and Ionic Equilibrium - Result Question 14
14. For the reaction, $SO _2(g)+\frac{1}{2} O _2(g) \rightleftharpoons SO _3(g)$ if $K _p=K _C(R T)^{x}$ where, the symbols have usual meaning, then the value of $x$ is (assuming ideality)
(2014 Main)
(a) $-1$
(b) $-\frac{1}{2}$
(c) $\frac{1}{2}$
(d) $1$
Show Answer
Answer:
Correct Answer: 14. (b)
Solution:
- For the given reaction, $\Delta n _g=n _P-n _R$
where, $n _P=$ number of moles of products
$ n _R=\text { number of moles of reactants } $
$ \begin{aligned} K _p & =K _c(R T)^{\Delta n _g} \\ \Delta n _g & =-\frac{1}{2} \end{aligned} $





