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3 years ago

Use Hess's Law to calculate the enthalpy change for the reaction

Chemistry

1 answer:

Marysya12 [62]3 years ago

3 0

Answer:

ΔH = 125.94kJ

Explanation:

It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:

1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ

2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ

-1/2 (1):

WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ

3/2 (2):

3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ

The sum of last both reactions:

WO3(s) + 3H2(g) → W(s) + 3H2O(g)

ΔH = 842.7kJ -716.76kJ

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What is the effect of high activation energy on a chemical reaction?

alex41 [277]

<h3><u>Answer;</u></h3>

It makes the reaction harder to start

<h3><u>Explanation</u>;</h3>

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The source of activation energy needed to push chemical reactions forward is obtained from the surroundings. Catalyst speed up chemical reaction by lowering the activation energy. Therefore, catalysis is the increase in the rate of a chemical reaction by lowering its activation energy.

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3 years ago

Boiling refers to which phase change?

Monica [59]

Boiling or also called evaporation is the conversion of liquid to gas through the application of heat. This phase change is an endothermic change and is the opposite of condensation from gas to liquid.

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2 years ago

Which ionic equation describes a redox reaction? A. Ag(+) + Cl- = AgCl B. 2H(+) + CO3(2-) = CO2 + H2O C. H(+) + OH(-) = H2O D. Z

Gennadij [26K]

<u>Answer:</u> The correct option is D. $Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

<u>Explanation:</u>

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously. It is known as the reaction in which the exchange of electrons takes place.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

From the given ionic reactions:

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

<u>On the reactant side: </u>

Oxidation number of Zn = 0

Oxidation number of Cu = +2

<u>On the product side: </u>

Oxidation number of Cu = 0

Oxidation number of Zn = +2

As the oxidation number of Zn is increasing from 0 to +2. Thus, it is getting oxidized. Similarly, the oxidation number of Cu is decreasing from +2 to 0. Thus, it is getting reduced. Therefore, forming a redox couple

Hence, the correct option is D. $Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

6 0

2 years ago

What is the volume of 300. g of mercury vapor at 822K and 0.900 atm?

Dominik [7]

Answer:

112.2L

Explanation:

Volume (V) = 300g

Temperature (T) = 822K

Pressure (P) = 0.9atm

using the ideal gas equation;

$PV = nRT\\\\ V = \frac{nRT}{P}$

Molar gas constant (R) = $0.0821L.atm/mol.K$

Mole (n) = $\frac{Mass (m)}{Molar mass (M)}$ Molar mass of Mercury = 200.59g/mol

$n = \frac{300g}{200.59 g/mol} \\$

= 1.496mol

Now, the volume can be calculated;

V = $\frac{1.496mol* 0.0821L.atm/mol.K*822K}{0.9atm}$

∴Volume of mercury = $112.2L$

8 0

2 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

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