`H_(2)(g) + I_(2)(g)hArr2HI(g)` When 46 g of `I_(2)` and 1 g of `H_(2)` gas heated at equilibrium at `450^(@)C`, the equilibrium mixture contained `1.9` g of `I_(2)` . How many moles of `I_(2)` and HI are present at equilibrium ?A. `0.0075` and `0.147` molesB. `0.005` and `0.147` molesC. `0.0075` and `0.347` molesD. `0.0052` and `0.347` moles

`H_(2)(g) + I_(2)(g)hArr2HI(g)` When 46 g of `I_(2)` and 1 g of `H_(2)

1.	`H_(2)(g) + I_(2)(g)hArr2HI(g)` When 46 g of `I_(2)` and 1 g of `H_(2)` gas heated at equilibrium at `450^(@)C`, the equilibrium mixture contained `1.9` g of `I_(2)` . How many moles of `I_(2)` and HI are present at equilibrium ?A. `0.0075` and `0.147` molesB. `0.005` and `0.147` molesC. `0.0075` and `0.347` molesD. `0.0052` and `0.347` moles
Answer» Correct Answer - C Moles of `I_(2)` taken `=(46)/(254) =0.181` Moles of `H_(2)` taken `=(1)/(2) =0.5` Moles of `I_(2)` remaining `=(1.9)/(254)=0.0075` Mols of `I_(2)` used `=0.81 - 0.0075 = 0.1735` Moles of `H_(2)` used `=0.1735` Moles of `H_(2)` reamining `=0.5 - 0.1735 = 0.3265` Moles of HI formed `=0.1735xx2=0.347` At equilibrium, moles of `I_(2) =0.0075` moles Moles of HI `=0.347` moles

Discussion

No Comment Found

Related InterviewSolutions

The vapour pressure of a liquid in a closed container depends uponA. 1 onlyB. 2 onlyC. 1 and 3 onlyD. 1,2,and3
Solubility of a solute in water is dependent on temperature as given by `S=Ae^(-DeltaH//RT)`, where `DeltaH`=heat of solution `"Solute"+H_(2)O(l) hArr "Solution", DeltaH= +- x` For given solution, variation of log S with temperature is shown graphically. Hence, solution is A. `CuSO_(4).5H_(2)O`B. NaClC. SucroseD. CaO
The `K_(c)` for `H_(2(g)) + I_(2(g))hArr2HI_(g)` is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will beA. `16`B. `32`C. `64`D. `128`
The `K_(c)` for `H_(2(g)) + I_(2(g))hArr2HI_(g)` is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will beA. 16B. 32C. 64D. 128
Following gaseous reaction is undergoing in a vessel `C_(2)H_(4) + H_(2)hArrC_(2)H_(6)` , `Delta H = - 32.7` Kcal Which will increase the equilibrium concentration of `C_(2)H_(6)` ?A. Increase of temperatureB. By reducing temperatureC. By removing some hydrogenD. By adding some `C_(2)H_(6)`
For the gas phase reaction `C_(2)H_(4)+H_(2) hArr C_(2)H_(6)(DeltaH=-32.7 "kcal")` carried out in a vessel, the equilibrium concentration of `C_(2)H_(4)` can be increased byA. Increasing the temperatureB. increasing concentration of `H_(2)`C. decreasing temperatureD. increasing pressure
The synthesis of ammonia is given as: `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g), DeltaH^(ɵ)=-92.6 kJ mol^(-1)` given `K_(c)=1.2` and temperature `(T)=375^(@)C` On increasing the temperature, the value of equilibrium constant `K_(c)`A. IncreasesB. DecreasesC. Remain unchangedD. Cannot be predicted
In an equilibrium `A+B hArr C+D`, A and B are mixed in vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the equilibrium has reaches, concentration of C was thrice the equilibrium concentration of B. Calculate `K_(c)`.
For the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g),DeltaH=-93.6 kJ mol^(-1)`. The concentration of `H_(2)` at equilibrium will increase ifA. the temperature is loweredB. the volume of the system is decreasedC. `N_(2)` is added at constant volumeD. `NH_(3)` is added
A tenfold increase in pressure on the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` at equilibrium result in ……….. in `K_(p)`.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment