Under certain conditions the rate of this reaction is zero order in hydrogen iodide with a rate constant of ·0.0045Ms−1:→2HIg+H2gI2gSuppose a 450.mL flask is charged under these conditions with 300.mmol of hydrogen iodide. How much is left 40.s later? You may assume no other reaction is important.Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Question
Under certain conditions the rate of this reaction is zero order in hydrogen iodide with a rate constant of ·0.0045Ms−1:→2HIg+H2gI2gSuppose a 450.mL flask is charged under these conditions with 300.mmol of hydrogen iodide. How much is left 40.s later? You may assume no other reaction is important.Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Solution
The reaction is zero order in hydrogen iodide, which means the rate of the reaction is independent of the concentration of hydrogen iodide. The rate equation for a zero order reaction is:
Rate = k
where k is the rate constant. In this case, k = 0.0045 M/s.
The integrated rate law for a zero order reaction is:
[HI]t = -kt + [HI]0
where [HI]t is the concentration of hydrogen iodide at time t, [HI]0 is the initial concentration of hydrogen iodide, and k is the rate constant.
First, we need to convert the amount of hydrogen iodide from mmol to mol, and then find the initial concentration in M (mol/L):
300 mmol = 0.3 mol
[HI]0 = 0.3 mol / 0.450 L = 0.667 M
Next, we can use the integrated rate law to find the concentration of hydrogen iodide after 40 seconds:
[HI]40 = -0.0045 M/s * 40 s + 0.667 M = 0.667 M - 0.18 M = 0.487 M
Finally, we convert this concentration back to mmol and round to 2 significant digits:
0.487 M * 0.450 L = 0.219 mol = 219 mmol
So, there are approximately 220 mmol of hydrogen iodide left after 40 seconds.
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