A sample of mixture of `A(g), B (g)` and `C(g)` under equilibrium has mean molecular mass equal to 80. The equilibrium is `:` `A(g)hArr B(g)+C(g)` If initially 4 mole of `'A'` gas is present then total number of mole at equilibrium is `:` `[M_(A)=100,M_(B)=60,M_(C)=40]`
A
5
B
2
C
6
D
4
Text Solution
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The correct Answer is:
1
`{:(A,hArr,B,+,C),(4-alpha,,4alpha,,4alpha):} ` `M_(av)=(M_(A))/(1+(2-1)alpha)n=(2)/(1)=2` `80=(100)/(1+(2-1)alpha)` `1+alpha=(100)/(80)` `prop(20)/(80)=(1)/(4)=0.25` total moles `=4+4alpha=5`
The equilibrium A(g) +4B(g) hArr AB_4(g) is attained by mixing equal moles of A and B in a one litre vessel Then at equilibrium
For the equilibrium DeltaB(g) hArr A(g) +B(g) K_p is equal to four times the total pressure. Calculate the number of moles of A formed .
The equilibrium constant K_(c) for A(g) hArr B(g) is 1.1 . Which gas has a molar concentration greater than 1 .
A(g) + 3B(g) rarr 4C(g) Initially concentration of A is equal to that of B. The equilibrium concentrations of A and C are equal. Kc is :
The equilibrium: P_(4)(g)+6Cl_(2)(g) hArr 4PCl_(3)(g) is attained by mixing equal moles of P_(4) and Cl_(2) in an evacuated vessel. Then at equilibrium:
K_p for the equation A(s)⇋ B(g) + C(g) + D(s) is 9 atm^2 . Then the total pressure at equilibrium will be
