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ZanzabumX [31]
3 years ago
14

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of t

his mixture releases 912 J of energy, to what volume will the gases expand against a constant pressure of 635 torr if all the energy of combustion is converted into work to push back the piston?
Chemistry
1 answer:
maks197457 [2]3 years ago
8 0

Answer:

The gases will expand until a volume of Vf= 10812 cm^3

Explanation:

Since the gas mixture is expanding at constant pressure

W=\int\limits^{Vf}_{Vi} {P} \, dV =P(Vf-Vi)

and therefore

Vf=W/P+Vi

knowing that 1 torr=133,32 Pa

p=635 torr * 133,322 Pa/torr * 1m^{3}/10^{6}cm^{3}=0,08466 J/ cm^{3}

therefore

Vf=912 J/(0,08466 J/cm^{3}) +40cm^{3} =10812 cm^{3}

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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

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Let the balanced reaction be ;

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Rate(R) of the reaction in trail 1 ,when :

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R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

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R=2.7\times 10^{-4} M/s

2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c...[2]

Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c...[3]

Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1

k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

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