20 gm of CaCO_(3)is allowed to dissociate in a 5.6 litres container at 819^(@) C. If 50% of CaCO_(3) is dissocitated at equilibrium the 'K_(p)' value is

20 gm of CaCO_(3)is allowed to dissociate in a 5.6 litres container at

1.	20 gm of CaCO_(3)is allowed to dissociate in a 5.6 litres container at 819^(@) C. If 50% of CaCO_(3) is dissocitated at equilibrium the 'K_(p)' value is
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>>5 atm<br/>1.6 atm<br/>4.8 atm<br/>10 atm</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E01_052_S01.png" width="80%"/> <br/> Molar con. at <a href="https://interviewquestions.tuteehub.com/tag/equlibrium-452655" style="font-weight:bold;" target="_blank" title="Click to know more about EQULIBRIUM">EQULIBRIUM</a> - `(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)/(5.6)` <br/> Given x=50% of 0.2 =0.1, `K_(c)=(CO_(2)]=(x)/(5.6)=(0.1)/(5.6)` <br/> but `K_(P)=K_(c)(RT)^(Delta <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>)=(0.1)/(5.6) xx (0.0821 xx 1092)^(1)=1.6` atm</body></html>