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Ratling [72]
2 years ago
6

A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and

measures the initial reaction rates (The data from the three experiments is in the table). 1. Write the rate law 2. Solve for k.

Chemistry
1 answer:
Dmitry_Shevchenko [17]2 years ago
6 0

Answer:

1. Rate =k [NO]^{2}[Cl_{2}]

2. k= 0.42 \frac{L^{2}}{mol^{2}*s}

Explanation:

Rate =k [NO]^{m}[Cl_{2}]^{n}

Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2

Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1

Rate =k [NO]^{2}[Cl_{2}]^{1}

Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12}  = 0.42 \frac{L^{2}}{mol^{2}*s}

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Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

Explanation:

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