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3 years ago

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What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio

n.Your answer should have two significant figures.Use −890.8kJmol for the molar heat of combustion of methane.

1 answer:

Veseljchak [2.6K]3 years ago

7 0

Answer:

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

$CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol$

Mass of methane burnt = 530 g

Moles of methane burnt = $\frac{530 g}{16 g/mol}=33.125 mol$

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

$-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ$

$-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ$

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

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Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

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Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

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Answer:

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Explanation:

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through the available space, colliding with one another and with the wall of container.

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<h3>Further explanation</h3>

Given

7.93 mol of dinitrogen pentoxide

T = 48 + 273 = 321 K

P = 125 kPa = 1,23365 atm

Required

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Solution

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From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :

= 0.5 x mol N₂O₅

= 0.5 x 7.93

= 3.965 moles

The volume of O₂ :

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