For the reaction H2(g) + I2(g) <===> 2HI(g), Kc=54.9 at 699K. A system was charged with 2.50 moles of H2 ans 2.50 moles of I2 and 5.00 moles of HI initially in a 5.00 liter vessell. How many moles of H2 should there be present at equilibrium?
(Answer my teacher gave: 0.2127mol in 1L)
My ICE TABLE
--H2(g) + I2(g) <===> 2HI(g)
I-(0.500)---(0.500)-----(1.00)
C-(-x)------(-x)--------(2x)
E-(0.500-x)-(0.500-x)---(1.00+2x)
54.9=(1.00+2x)^2/(0.500-x)^2
54.9=4x^2+4x+1 / x^2-x+0.25
54.9x^2-54.9x+13.725=4x^2+4x+1
50.9x^2-58.9x+12.725=0
Then I used the quadratic formula to get my answer
(using + root)x=0.8697
(uding - root)x=0.2874
0.500-0.2874=0.2126
My Issue:
I get the correct answer but I get the answer using the using root. I thought this should never happen. I also know another method to solve this question and when I do it I get x=0.2874.
Well, you should have proofed your post. The first sentence at the end of your post says "I get the correctanswer but I the answers using the USING ROOT".I don't know what a using root is I assume you meant to say you get the correct answer using the negative root and .... The 0.2874 is the correct answer for x(based on your numbers--I didn't solve the quadratic) BECAUSE it gives you 0.500 - 0.2874 = 0.2126 and that is a reasonable answer for the H2 and I2 after equilibrium is attained. What do you get for H2 and I2 if you use the + root of 0.8697? That would be 0.5-0.8697 and that is a negative concentration for H2 and I2 and you know that can't be right. #1 ruleis you shouldn't worry about it because you KNOW which is the right answer and #2 I would forget about that rule about the negative root NEVER being right. In this case the FINAL answer is what counts and not an intermediate result In fact, I
NEVER pay any attention to which root is being used because it doesn't matter.
The method you used to solve the quadratic equation is correct, and you have obtained the correct roots. It's important to note that both roots are valid solutions in this case.
In the ICE table, the change in moles of H2 is represented by -x. However, in the equilibrium expression, the value of x can be positive or negative. This is because the change in concentration can either be a decrease or an increase depending on the direction of the reaction.
In this particular case, the positive root (x = 0.8697) represents an increase in the concentration of H2 at equilibrium, while the negative root (x = 0.2874) represents a decrease in the concentration of H2 at equilibrium. Since we are given that there were initially 2.50 moles of H2, it is not possible for the concentration to decrease to a negative value. Thus, we consider the positive root as the correct answer.
Therefore, at equilibrium, the moles of H2 should be 2.50 - 0.8697 = 1.6303 moles.
The reason you obtained a different answer using the other method is possibly due to some calculation errors or a different approach. It's better to double-check your calculations and make sure you are following the correct steps.
I hope this helps clarify your doubt. Let me know if you have any further questions!
It seems like you have made a small mistake in your calculations. Let's go through your work to find the error.
You correctly set up the ICE table and calculated the equilibrium concentrations as:
H2: 0.500 - x
I2: 0.500 - x
HI: 1.00 + 2x
You then used the expression for the equilibrium constant, Kc, to set up the equation:
Kc = (1.00 + 2x)^2 / ((0.500 - x)^2) = 54.9
Simplifying this equation gives you:
(1.00 + 2x)^2 = 54.9 * (0.500 - x)^2
Expanding both sides of the equation gives you:
1.00 + 4x + 4x^2 = 54.9 * (0.250 - x + x^2)
Next, you made an error in your calculation. The correct equation should be:
4x^2 + 4x + 1 = 54.9 * (0.250 - x) + 54.9 * x^2
Now we can simplify the equation:
4x^2 + 4x + 1 = 13.725 - 54.9x + 54.9x^2
Combining like terms, we get:
50.9x^2 - 58.9x + 12.725 = 0
Now we can use the quadratic formula to find the value of x:
x = (-b ± √(b^2 - 4ac))/(2a)
Applying this formula to the equation, we get:
x = (-(-58.9) ± √((-58.9)^2 - 4(50.9)(12.725)))/(2(50.9))
Calculating this expression gives us two possible values for x:
x ≈ 0.2127 or x ≈ 0.2874
From your teacher's answer, it seems that they used the positive root, x ≈ 0.2127. However, both values are valid solutions as they correspond to different points in the reaction process.
It's important to note that different methods of solving the equation may yield different values of x depending on the assumptions made or the approximations used. In this case, both values are correct, and your method is valid as well.
Therefore, it is possible to get the answer using either root, as long as the calculation is performed accurately.