5 moles of `SO_(2)`and 5 moles of `O_(2)` are allowed to react .At equilibrium , it was foumnd that `60%` of `SO_(2)` is used up .If the pressure of the equilibrium mixture is one aatmosphere, the parital pressure of `O_(2)` is :A. 0.82 atmB. 0.52 atmC. 0.21 atmD. 0.41 atm

5 moles of `SO_(2)`and 5 moles of `O_(2)` are allowed to react .At equ

1.	5 moles of `SO_(2)`and 5 moles of `O_(2)` are allowed to react .At equilibrium , it was foumnd that `60%` of `SO_(2)` is used up .If the pressure of the equilibrium mixture is one aatmosphere, the parital pressure of `O_(2)` is :A. 0.82 atmB. 0.52 atmC. 0.21 atmD. 0.41 atm
Answer» Correct Answer - D `{:(,2SO_(2)(g),+,O_(2)(g),hArr,2SO_(3)(g)),("Initially",5,,5,,0),("Used up",(5xx60)/(100)=3,,3xx(1)/(2)=1.5,,1.5xx2=3),("At eqm. ",5-3=2,,5-1.5=3.5,,):}` `:. p_(O_(2))=("Moles of" O_(2))/("Total no. of moles" )xx"Total pressure "` `=(3.5)/(2+3.5+3)xx1=(3.5)/(8.5)xx1=0.41 "atm."`

Discussion

No Comment Found

Related InterviewSolutions

The exothermic formation of `ClF_(3)` is represented by thr equation: `Cl_(2)(g)+3F_(2)(g) hArr 2ClF_(3)(g), DeltaH=-329 kJ` Which of the following will increase the quantity of `ClF_(3)` in an equilibrium mixture of `Cl_(2), F_(2)`, and `ClF_(3)`?A. increasing the temperatureB. removing `Cl_(2)`C. increasing the volume of the containerD. adding `F_(2)`
The equilibrium constant for the reaction `CaSO_(4).5H_(2)H_(2)O(s)hArr CaSO_(4).3H_(2)O(s)+2H_(2)O(g)` is equal toA. `([CaSO_(4).3H_(2)O][H_(2)O]^(2))/([CaSO_(4).5H_(2)O])`B. `([CaSO_(4).3H_(2)O])/([CaSO_(4).5H_(2)O])`C. `[H_(2)O]^(2)`D. `[H_(2)O].`
Which of the following equilibrium will shift left on increasing temperature. `(i) " " H_(2)O(g) hArr +1//2 O_(2)(g)` `(ii) " " CO_(2)(g) +C(s) hArr CO(g)` `(iii) " " C(s, "diamond") hArr C(s, "graphite")`A. Only (i)B. Both (i) and (ii)C. Only (iii)D. None
Consider the following reversible reactionat equilibrium: `2H_(2)O(g) hArr 2H_(2)(g)+O_(2)(g), DeltaH=+24.7 kJ` Which one of the following changes in conditions will lead to maximum decomposition of `H_(2)O(g)`?A. Increasing both temeprature and pressureB. Decreasing temperature and increasing pressureC. Increasing temperature and decreasing pressureD. Increasing temperature at constant pressure
Given the equilibrium system `:` `NH_(4)Cl(g) hArr NH_(44)^(+)(aq)+Cl^(-)(aq), (DeltaH = + 35 kcal //` mol ). What change will shift the equilibrium to the right ?A. decreasing the temperatureB. increasing the temperatureC. dissolving NaCl crystals in the equilibrium mixtureD. dissolving `NH_(4)NO_(3)` crystals in the equlibrium mixture
There is 50% dimer formation of benzoic acid `(C_(6)H_(5)COOH)_(2)` in benzene solution. `2C_(6)H_(5)COOHhArr (C_(6)H_(5)COOH)_(2)`. Hence abnormal molecular weight of benzoic acid (theorectical value `=122g mol^(-1))` isA. 61 g `mol^(-1)`B. 244 g `mol^(-1)`C. 163 g `mol^(-1)`D. 81 g `mol^(-1)`
At equilibrium of the reaction , `N_(2)O_(4)(g) hArr 2NO_(2)(g)` the observed molecular weight of `N_92)O_(4)` is 80g `mol^(-1)` at 350K. The percentage dissociation of `N_(2)O_(4)(g)` at 350K isA. `10%`B. `15%`C. `20%`D. `18%`
Before equilibrium is set-up for the chemical reaction, `N_(2)O_(4)hArr 2NO_(2)`, vapour density of the gaseous mixture was measured. If D is the theoretical value of vapour density, variation of x with `D//d` is by the graph.
In the dissociation of `N_(2)O_(4)` into `NO_(2). (1+x)` values with the vapour densities ratio `((D)/(d))` is given by `:`A. B. C. D.
The vapour density of `N_(2)O_(4)` at a certain temperature is `30`. Calculate the percentage dissociation of `N_(2)O_(4)` this temperature.A. `53.3%`B. `106.6%`C. `26.7%`D. None

5 moles of `SO_(2)`and 5 moles of `O_(2)` are allowed to react .At equilibrium , it was foumnd that `60%` of `SO_(2)` is used up .If the pressure of the equilibrium mixture is one aatmosphere, the parital pressure of `O_(2)` is :A. 0.82 atmB. 0.52 atmC. 0.21 atmD. 0.41 atm

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment