Once two gas particles undergo an elastic collision, the particles

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

devlian [24]

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

6 0

4 years ago

Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca

lina2011 [118]

<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

$Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)$

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.23-0.77=-1.0V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

3 0

3 years ago

If I initially have a gas at a pressure of 12 kPa, a volume of 23 liters, and a temperature of 200 K, and then I raise the press

sertanlavr [38]

Answer:

The new volume of gas would be 30 L.

Explanation:

This is an example of a Combined Gas Laws problem.

5 0

3 years ago

How can glass be a liquid if it's so hard?

levacccp [35]

Answer:

It is an amorphous solid and hence also called pseudo solid. So it flows very slowly over thousands of years. It is not visible to the n*ked eye.

7 0

2 years ago

What is liquor ammonia fortis?

erik [133]

Answer:

Ammonia fortis liquor is a saturated solution of ammonia in water. It is also called 880 ammonia. Its relative density is 0.880. It is stored in tightly sealed bottles in a cold place. (Sorry if I'm wrong)

Explanation:

6 0

3 years ago

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