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In the reaction `COCl_2(g)hArr CO(g)+Cl_2(g)` at `550^@C`, when the initial pressure of CO & `Cl_2` are 250 and 280 mm of Hg respectively. The equilibrium pressure is found to be 380 mm of Hg.Calculate the degree of dissociation of `COCl_2` at 1 atm.What will be the extent of dissociation, when `N_2` at a pressure of 0.4 atm is present and the total pressure is 1 atm.A. 0.32 and no changeB. 0.32 and 0.4C. 0.4 and 0.3D. In presence of `N_2` dissociation cannot take place |
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Answer» Correct Answer - B `{:(,CoCl_2(g)hArr,CO(g)+Cl_2(g)),("Initial pressure",-,250" "280),("Equilibrium pressure",x,250-x " " 280-x):}` x+250-x+280-x=380 x=150 `K_P=0.114` `K_P=(P_(CO).P_(Cl_2))/(P_(COCl_2))implies K_P=(alpha^2 . P)/(1-alpha^2) implies 0.114 =(alpha^2 . 1)/(1-alpha^2)` hence `alpha=0.32 " " alpha=0.32` In presence of `N_2` (constant pressure process) `K_P=(alpha^2xx0.6)/(1-alpha^2)implies alpha=sqrt(0.114/0.714)implies alpha=0.4` `alpha` increases from 0.32 to 0.4 |
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