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  • For the reaction :             <img alt="\mathrm{2HI(g) \rightleftharpoons H_2(g) + I_2 (g)}" src="https://entrancecorner.oncod

For the reaction :

            \mathrm{2HI(g) \rightleftharpoons H_2(g) + I_2 (g)}

The value of the degree of dissociation in terms of the equilibrium constant KP is given in 

Option: 1

\mathrm{\frac{2\sqrt {K_P}}{1+ 2 \sqrt{K_P}}}


Option: 2

\mathrm{\frac{\sqrt {2K_P}}{1+ \sqrt{2K_P}}}


Option: 3

\mathrm{\frac{4\sqrt {K_P}}{1+ 5 \sqrt{K_P}}}


Option: 4

\mathrm{\sqrt {\frac{2\ {K_P}}{1+ 2\ {K_P}}}}


Answers (1)

best_answer

                \mathrm{2HI(g) \rightleftharpoons H_2(g) + I_2 (g)}

t = 0             \text{a}                   \text{0}                  \text{0}

t = teq    \mathrm{a(1-\alpha)}         \mathrm{\frac{a\alpha}{2}}             \mathrm{\frac{a\alpha}{2}}

\mathrm{\therefore K_P = \frac{ \frac{a\alpha}{2} \times \frac{a\alpha}{2} }{a^2(1-\alpha)^2}= \frac{\alpha^2}{4(1-\alpha)^2}}

\mathrm{\therefore \frac{\alpha^2}{(1-\alpha)^2}=4K_P}

\mathrm{\Rightarrow \frac{\alpha}{1-\alpha}=2\sqrt{K_P}}

\mathrm{\Rightarrow \frac{\alpha}{1-\alpha+\alpha}=\frac{2\sqrt{K_P}}{1 +2\sqrt{K_P}}}

\mathrm{\Rightarrow \alpha=\frac{2\sqrt{K_P}}{1 +2\sqrt{K_P}}}

Hence, the correct answer is Option (1)

Posted by

Sanket Gandhi

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