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Naya [18.7K]
4 years ago
7

Q4: The average heat evolved by the oxidation of food in an average adult per hour per kilogram of body weight is 7.20 ୩୎ ୩୥ ୦୰.

Assume the weight of an average adult is 62.0 kkg. Suppose the total heat evolved by this oxidation is tranderred into the surroundings over a period of 1 week. Calculate the entropy change of the surroundings associated with this heat transfer. Assume the surroundings are at 293K.
Chemistry
1 answer:
8_murik_8 [283]4 years ago
4 0

Answer:

The entropy change of the surroundings associated with this heat transfer is 255.9J/K

Explanation:

The heat evolved by the oxidation of food is 7.2J/Kg.h thus, multiplying the heat by the weight of the person and by the time the total heat (q) is obtained. In this case, the weight is 62.0Kg and a weak has 168 hours.

q = 7.2\frac{J}{Kg.h}x62.0Kgx168h = 74995.2J

The entropy change of the surrounding associated with this heat transfer can be calculated by the ratio between the heat exchanged and the temperature.

ΔS = \frac{q}{T} = \frac{74995.2J}{293K} = 255.9J/K

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The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed
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Answer:

(a) The rate of disappearance of O_{2} is: 4.65*10^{-5} M/s

(b) The value of rate constant is: 0.83036 M^{-2}s^{-1}

(c) The units of rate constant is:  M^{-2}s^{-1}

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The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

2NO(g)+O_{2}->2NO_{2}

rate = -\frac{1}{2} \frac{d}{dt}[NO] = -\frac{d}{dt}[O_{2}] = \frac{1}{2}\frac{d}{dt}[NO_{2}] -----(1)

According to the question, the reaction is second order in NO and first order in  O_{2}.

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-\frac{d}{dt}[NO] = 9.3*10^{-5} M/s.

(a) From (1), we can get the rate of disappearance of O_{2}.

    Rate of disappearance of  O_{2} = -\frac{d}{dt}[O_{2}] = (0.5)*(9.3*10^{-5}) M/s = 4.65*10^{-5} M/s.

(b) The rate of the reaction can be obtained from (1).

    rate = -\frac{1}{2} \frac{d}{dt}[NO] = (0.5)*(9.3*10^{-5})

    rate = 4.65*10^{-5} M/s

   The value of rate constant can be obtained by using (2).

    rate constant = k = \frac{rate}{[NO]^{2}[O_{2}]}

    k = \frac{4.65*10^{-5}}{(0.040)^{2}(0.035)} = 0.83036 M^{-2}s^{-1}

(c) The units of the rate constant can be obtained from (2).

    k = \frac{rate}{[NO]^{2}[O_{2}]}

    Substituting the units of rate as M/s and concentrations as M, we get:

\frac{Ms^{-1} }{M^{3}} = M^{-2}s^{-1}

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

     rate\alpha [NO]^{2}

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of (1.8)^{2} = 3.24

     

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