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saul85 [17]
3 years ago
13

CO(g) + 12 O2(g) → CO2(g)The combustion of carbon monoxide is represented by the equation above.(a) Determine the value of the s

tandard enthalpy change, ∆HrxnD , for the combustion of CO(g) at 298 Kusing the following information. C(s) + 12 O2(g) → CO(g) ∆H298D = − 110.5 kJ mol−1 C(s) + O2(g) → CO2(g) ∆H298D = − 393.5 kJ mol−1
Chemistry
1 answer:
devlian [24]3 years ago
6 0

Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

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.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The combustion of CO will be,

CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)     \Delta H_1=-110.5kJ/mol

(2) C(s)+O_2(g)\rightarrow CO_2(g)     \Delta H_2=-393.5kJ/mol

Now we are reversing reaction 1 and then adding both the equations, we get :

(1) CO(g)\rightarrow C(s)+\frac{1}{2}O_2(g)     \Delta H_1=110.5kJ/mol

(2) C(s)+O_2(g)\rightarrow CO_2(g)     \Delta H_2=-393.5kJ/mol

The expression for enthalpy change for the reaction will be,

\Delta H_{rxn}=\Delta H_1+\Delta H_2

\Delta H_{rxn}=(110.5)+(-393.5)

\Delta H_{rxn}=-283kJ/mol

Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

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You have dissolved 10 g sodium oxide in 200 ml water.calculate concentration of the solution
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0.85 Molar Na2O

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Determine the moles of sodium oxide, Na2O, in 10 grams by dividing by the molar mass of Na2O (61.98 g/mole).

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Molar is a measure of concentration.  It is defined as moles/liter.  A 1 M  solution contains 1 mole of solute per liter of solvent.  [200 ml water = 0.2 Liters water.]

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3 years ago
Lead(II) oxide from an ore can be reduced to elemental lead by heating in a furnace with carbon. Calculate the expected yield of
Lelu [443]

Answer:

53kg is the expected yield of lead

Explanation:

Firstly, in order to solve this question, we need to write the equation of reaction correctly. This is as follows:

PbO(s) + C(s) ---> Pb(l) + CO(g)

We proceed from here. We should get the limiting reactant but this can only be obtained by getting the number of moles of each reactant present.

The formula to use across all boards is that the number of moles is the mass of each of the reactant divided by the molar mass of each of the reactant.

For PBO, mass is 57kg = 57000g

Molar mass of PBO = 223.20g/mol

The number of moles is thus 57,000/223.2 = 255.37 moles

For carbon, mass is also 57kg = 57000g

Molar mass is 12g/mol

Number of moles of carbon = 57000/12 = 4750 moles

From the number of moles, we can see that the number of moles of Carbon is greater than that of PbO. This means that PbO is the limiting reagent.

Hence we use it to calculate percentage yield.

The number of moles of lead formed is the same of number of moles of lead oxide = 255.37 since we have mole ratio of 1 to 1

The molar mass of lead is 207.20g/mol

The mass of lead formed is = moles of lead formed * molar mass of lead = 207.20 * 255.37 = 52,912g which is approximately 53kg

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6 0
3 years ago
A first-order reaction is 45% complete at the end of 43 minutes. What is the length of the half-life of this reaction
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The half-life of the reaction is 50 minutes

Data;

  • Time = 43 minutes
  • Type of reaction = first order
  • Amount of Completion = 45%

<h3>Reaction Constant</h3>

Let the initial concentration of the reaction be X_0

The reactant left = (1 - 0.45) X_0 = 0.55 X_0 = X

For a first order reaction

\ln(\frac{x}{x_o}) = -kt\\ k = \frac{1}{t}\ln (\frac{x_o}{x}) \\ k = \frac{1}{43}\ln (\frac{x_o}{0.55_o})\\ k = 0.013903 min^-^1

<h3>Half Life </h3>

The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.

x = \frac{x_o}{2} \\t = t_\frac{1}{2} \\t_\frac{1}{2} = \frac{1}{k}\ln(\frac{x_o}{x_o/2})\\

Substitute the values

t_\frac{1}{2} = \frac{1}{k}\ln(2)=\frac{0.6931}{0.013903}\\t_\frac{1}{2}= 49.85 min = 50 min

The half-life of the reaction is 50 minutes

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