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  •  of HI undergoes deomposition to <img alt="\mathrm{H}_{2}" src="https://entrancecorner.oncodecogs.com/

40 \% of HI undergoes deomposition to \mathrm{H}_{2} and \mathrm{I}_{2} \text { at } 300 \mathrm{~K} \cdot \Delta \mathrm{G}^{\ominus} for this decompostion reaction at one atmopsphere pressure is_____________ \mathrm{J} \mathrm{mol}^{-1}.[nearest integer]

The decomposition reaction is given as 

\mathrm{HI \rightleftharpoons \frac{1}{2}H_2+ \frac{1}{2}I_2}

\text { (Use } \mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \log 2=0.3010, \ln 10=2.3, \log 3=0.477 \text { ) }

Option: 1

2735


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\\\mathrm{H I \rightleftharpoons \frac{1}{2} H_{2}+\frac{1}{2} I_{2}}\\ \mathrm{P_{0}}

\mathrm{0.6 P_{0} \quad 0.2 P_{0} \quad 0.2 P_{0}}

Given,

\begin{aligned} &\mathrm{P_{T}=1=(0.6+0.2+0.2) P_{0}} \\ \\&\Rightarrow \mathrm{P_{0}=1} \end{aligned}

\mathrm{\therefore K_{P}=\frac{P_{H_{{2}}}^{1 / 2} P_{I_{{2}}}^{1 / 2}}{P_{H I}}=\frac{(0.2)}{(0.6)}=\frac{1}{3}}

\therefore \mathrm{\Delta G^{\circ}=-R T \ln K_P=+8.31 \times 300 \times 2.3 \times \log{3}}

                                            =2735 \text{ J mole}^{-1}

Hence, answer is 2735

Posted by

shivangi.shekhar

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