A(g) + B(g) <=> AB(g)
At a given temperature 1.0 mole of A and 1.0 mole of B are placed in the 1.0 litre vessel and allowed to reach equilibrium. Analysis revealed that the equilibrium concenration of AB was 0.40 molar. What percent of A had been converted to products?
I'm not really sure how to solve this problem. Do I find the equilibrium concentration of A and then divide the equilibrium cocentration A with equilibrium concentration AB, times 100?
A/AB*100 = ??
Please help. Thanks!
I think what you have written is the percent of A that remains. What you want is the percent converted, which is (AB/A)*100 = (0.40/1.0)*100 = ??
0.4 is the amount that was "removed" from A and that is what is converted. Check my thinking.
oh, so i don't need to find the equilibirum concentration of A? I just use the initial concentration given (1.0)?
That't right. Notice that the problem gives you a starting value for A (1.0 mol/L) AND an equilibrium value for AB(0.40) and asks for the percent that changed. Think of it as 10 apples and 10 oranges in a basket and we peel and slice 5 of the apples and 5 of the oranges to make a salad. We have used 50% of the apples. In the problem, 40% of the A molecules have been converted. We could also say that if 0.40 was converted, then 60% were not converted, and 100 - 60 = 40% converted.
Yes, that's correct! You don't need to find the equilibrium concentration of A because it is not directly given in the problem. Instead, you can use the initial concentration of A (1.0 mole/L) to calculate the percent of A that has been converted to products.
To solve the problem, you can use the formula: (AB/A) * 100 = (0.40/1.0) * 100 = 40%.
This calculation tells you that 40% of the initial concentration of A has been converted to form the product AB. Therefore, the percent of A that has been converted to products is 40%.
Remember to always check your calculation and make sure you are using the correct values and units in your formula.