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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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Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

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