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3 years ago

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5. For which of the following reactions would the heat of reaction be labelled ∆Hf°? a) PCl3 + ½ O2(g) ->POCl3(g) b) CaO(s) +

SO2(g) -> CaSO3(s) c) Al(s) + 3/2O2(g) + 3/2H2(g) -> Al(OH)3(s) d) ½ N2O(g) + 1/4 O2(g) -> NO(g)

1 answer:

user100 [1]3 years ago

8 0

Answer:

C

Explanation:

∆H°f means the enthalpy change of formation of one mole of substance by its constituent elements under standard conditions.

So in an equation for ∆H°f, we must see 2 or more elements as reactants combining to form a compound.

In the 4 answers, only C represents elements forming a substance (Al(s) + 3/2O2(g) + 3/2H2(g) -> Al(OH)3(s)),

while the others include compounds as one of their reactants.

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Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

<u>Answer:</u>

<u>For a:</u> The total heat required is 36621.5 J

<u>For b:</u> The total heat required is 58944.5 J

<u>Explanation:</u>

<u>For a:</u>

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

<u>For process 1:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

<u>For process 2:</u>

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

<u>For b:</u>

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

<u>For process 1:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

<u>For process 2:</u>

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

<u>For process 3:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

<u>For process 4:</u>

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

8 0

3 years ago

When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x

Leno4ka [110]

You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles

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2 years ago

How is ti possible that, when adding protons while moving left to right across a period, the size of an atom shrinks? Why is the

Kaylis [27]

As one moves across a period, from left to right, both the number of protons and electrons of a neutral atom increase. The enhancing number of electrons and protons results in a greater attraction between the electrons and the nucleus. This uplifted attraction pulls the electrons nearer to the nucleus, therefore, reducing the size of the atom.

On the other hand, while moving down a group, there is an increase in the number of energy levels. The enhanced number of energy levels increases the size of the atom in spite of the elevation in the number of protons. In the outermost energy levels, the protons get attracted towards the nucleus, however, the attraction is less due to an increase in the distance from the nucleus.

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3 years ago

What is the temperature when ice Melt

kenny6666 [7]

The temperature when ice melts is 32 degrees Fahrenheit.

6 0

3 years ago

Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d

worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B\rightarrow Products$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

$3.20\times 10^{-1}=k(1.50)^a(1.50)^b$ ....(1)

Expression for rate law for second observation:

$3.20\times 10^{-1}=k(1.50)^a(2.50)^b$ ....(2)

Expression for rate law for third observation:

$6.40\times 10^{-1}=k(3.00)^a(1.50)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0$

Dividing 1 from 3, we get:

$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[A]^1[B]^0$

$\text{Rate}=k[A]$

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

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