Following reaction is given CH 3 COCH 3( g ) leftharpoons CH 3- CH 3+ CO ( g ), initial pressure of CH 3 COCH 3 is 100 mm of Hg. When equilibrium is set up, the mole fraction of CO ( g ) is 1 / 3, hence K p is-

Question

Following reaction is given CH 3 COCH 3( g ) leftharpoons CH 3- CH 3+ CO ( g ), initial pressure of CH 3 COCH 3 is 100 mm of Hg. When equilibrium is set up, the mole fraction of CO ( g ) is 1 / 3, hence K p is- (A) 10 mm (B) 50 mm (C) 25 mm (D) 150 mm

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/fc9c074e159e3445a4d07d603f66a54a-.png" /> Since, rest all parametess are constant (v, T), we can say that PC O(at eqm)=(1/3) P text Total (at eqm) P=(1/3)(100-P+P+P) ⇒ P=(1/3)(100+P) 3 p=100+p ⇒ 2 p=100 p=50 mm of Hg K p =(( pCH 3- CH 3)( pCO )/( pCH 3 COCH 3))=(50 × 50/(100-50))=(50 × 50/50) =50 mm

Q. Following reaction is given CH3​COCH3​(g)⇌CH3​−CH3​+CO(g), initial pressure of CH3​COCH3​ is 100mm of Hg. When equilibrium is set up, the mole fraction of CO(g)​ is 1/3, hence Kp​ is-

10 mm

50 mm

25 mm

150 mm

Since, rest all parametess are constant (v, T), we can say that PCO(ateqm)​​=31​PTotal (ateqm)​​ P=31​(100−P+P+P) ⇒P=31​(100+P) 3p=100+p ⇒2p=100 p=50mm of Hg Kp​=(pCH3​COCH3​)(pCH3​−CH3​)(pCO)​=(100−50)50×50​=5050×50​ =50mm

Q. Following reaction is given $C H_{3} COC H_{3} (g) ⇌ C H_{3} - C H_{3} + CO (g),$ initial pressure of $C H_{3} COC H_{3}$ is $100 mm$ of $H g$ . When equilibrium is set up, the mole fraction of $C O_{(g)}$ is $1/3,$ hence $K_{p}$ is-