A mixture of one mole of CO_(2) and one mole of H_(2) attains equilibrium at a temperature of 250^(@)C and a total pressure of 0.1 atm for the change CO_(2(g)) +H_(2(g)) harr CO_((g))+H_(2)O_((g)). Calculate K_(P) if the analysis of final reaction mixture shows 0.16 volume fraction of CO

A mixture of one mole of CO_(2) and one mole of H_(2) attains equilibr

1.	A mixture of one mole of CO_(2) and one mole of H_(2) attains equilibrium at a temperature of 250^(@)C and a total pressure of 0.1 atm for the change CO_(2(g)) +H_(2(g)) harr CO_((g))+H_(2)O_((g)). Calculate K_(P) if the analysis of final reaction mixture shows 0.16 volume fraction of CO
Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>0.46<br/>0.63<br/>0.22<br/>0.82</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E02_034_S01.png" width="80%"/> <br/> volume <a href="https://interviewquestions.tuteehub.com/tag/fraction-458259" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTION">FRACTION</a> = pressure fraction = mole fraction <br/> `X_(<a href="https://interviewquestions.tuteehub.com/tag/co-920124" style="font-weight:bold;" target="_blank" title="Click to know more about CO">CO</a>)=X_(H_(2)O)=0.16` <br/> `X_(CO_(2))=X_(H_(2))=(1-2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.16)/(2)=0.34` <br/> `K_(P)=((X_(CO) xx P)(X_(H_(2)O) xx P))/((X_(CO_(2)) xx P)(X_(H_(2)O) xx P))` <br/> `=(0.16 xx 0.16)/(0.34 xx 0.34)=0.22`</body></html>