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Anna35 [415]
3 years ago
13

The first-order rate constant for the decomposition of n2o5, 2n2o5(g)→4no2(g)+o2(g) at 70∘c is 6.82×10−3 s−1. suppose we start w

ith 2.00×10−2 mol of n2o5(g) in a volume of 2.3 l . you may want to reference (page) section 14.4 while completing this problem. part a how many moles of n2o5 will remain after 4.0 min ?,
Chemistry
1 answer:
torisob [31]3 years ago
7 0

The rate constant for 1st order reaction is

K = (2.303 /t) log (A0 /A)

Where, k is rate constant

t is time in sec

A0 is initial concentration

(6.82 * 10-3) * 240 = log (0.02 /A)

1.63 = log (0.02 /A)

-1.69 – log A = 1.63

Log A = - 0.069

A = 0.82

Hence, 0.82 mol of A remain after 4 minutes.

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Molarity after dilution : 0.0058 M

<h3>Further explanation </h3>

The number of moles before and after dilution is the same  

The dilution formula

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V₁=6.11

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\tt M_2=\dfrac{M_1.V_1}{V_2}=\dfrac{0.1\times 6.11}{105.12}=0.0058~M

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