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sleet_krkn [62]
3 years ago
7

The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentrati

on of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?
Chemistry
1 answer:
Gennadij [26K]3 years ago
5 0

This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.42\times 10^{-2}\text{ hr}^{-1}

t = time passed by the sample  = ?

a = initial concentration of the reactant  = 0.080 M

a - x = concentration left = 0.053 M

Now put all the given values in above equation, we get

t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}

t=17.0\text{ hr}

Therefore, the time taken will be, 17.0 hr

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Answer:

The equilibrium concentration of I₂ is 0.400 M  and HI is 1.20 M, the Keq will be 0.112.

Explanation:

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It is given that 4.00 mol of HI was filled in a flask of 2.00 L, thus, the concentration of HI will be,  

= 4.00 mol/2.00 L

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Based on the reaction, the initial concentration of 2HI is 2.00, H₂ is 0 and I₂ is O. The change in the concentration of 2HI is -x, H₂ is x and I₂ is x. The equilibrium concentration of 2HI will be 0.200-x, H₂ is x and I₂ is x.  

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Now the equilibrium concentration of HI will be,  

= 2.00 -2x  

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