Answer
410.1k+ views
Hint: In this we have to check all the given thermodynamic equations by making one equation from the other two given equations and so on and then we can easily find the correct thermodynamic equations for the given reactions.
Complete step by step answer:
By the term the enthalpy of formation, we simply mean the total change in the enthalpy of the reaction when 1mole of the compound is formed from its constituents’ elements and it is represented by $\Delta {{H}^{\circ }}$.
To know, which thermodynamic relation is correct, we will check the all given equations one by one as;
(a) z= x+ y
Consider the z equation:
$CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$---------(1)
We have to make this equation from the rest two equations;
The rest two equation are as;
$C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=x\text{ }kJ\text{ }mol{{e}^{-1}}$----------(2)
$C(graphite)+\dfrac{1}{2}{{O}_{2}}(g)\to CO(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=y\text{ }kJ\text{ }mol{{e}^{-1}}$----------(3)
We can see that , in equation (1), CO is on the reactant side, we so will invert equation (3) so that the CO will come on reactant side and now it ${{\Delta }_{r}}{{H}^{\circ }}=-y\text{ }kJ\text{ }mol{{e}^{-1}}$. And add equations (2) and (3) to get equation (1), then;
z= x-y
so , option(a) is incorrect.
(b) x=y-z
Consider the x equation:
$C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=x\text{ }kJ\text{ }mol{{e}^{-1}}$----------(4)
We have to make this equation from the rest two equations;
The rest two equation are as;
$CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$----------(5)
$C(graphite)+\dfrac{1}{2}{{O}_{2}}(g)\to CO(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=y\text{ }kJ\text{ }mol{{e}^{-1}}$----------(6)
On adding the equations (5) and (6), we will get the equation(4), then;
x=y +z
so, option(b) is also incorrect.
(c) x=y+ z
From the option (b), we come to know that the option (c) is correct.
(d) y=2z-x
Consider the y equation :
$C(graphite)+\dfrac{1}{2}{{O}_{2}}(g)\to CO(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=y\text{ }kJ\text{ }mol{{e}^{-1}}$---------(7)
We have to make this equation from the rest two equations;
The rest two equation are as;
$CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$----------(8)
$C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=x\text{ }kJ\text{ }mol{{e}^{-1}}$----------(9)
We can see that , in equation (7), CO is on the product side, we so will invert equation (8) so that the CO will come on product side and now it ${{\Delta }_{r}}{{H}^{\circ }}=-z\text{ }kJ\text{ }mol{{e}^{-1}}$. And add equations (8) and (9) to get equation (7), then;
y= x-z
so, option(d) is also incorrect.
Hence, from the above we can see that the correct option is (c).
Note: The enthalpy of formation of elements which are present in their molecular forms like oxygen gas, or in any solid form etc. their standard enthalpy of formation is always taken as zero as they undergo no change in their formation.
Complete step by step answer:
By the term the enthalpy of formation, we simply mean the total change in the enthalpy of the reaction when 1mole of the compound is formed from its constituents’ elements and it is represented by $\Delta {{H}^{\circ }}$.
To know, which thermodynamic relation is correct, we will check the all given equations one by one as;
(a) z= x+ y
Consider the z equation:
$CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$---------(1)
We have to make this equation from the rest two equations;
The rest two equation are as;
$C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=x\text{ }kJ\text{ }mol{{e}^{-1}}$----------(2)
$C(graphite)+\dfrac{1}{2}{{O}_{2}}(g)\to CO(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=y\text{ }kJ\text{ }mol{{e}^{-1}}$----------(3)
We can see that , in equation (1), CO is on the reactant side, we so will invert equation (3) so that the CO will come on reactant side and now it ${{\Delta }_{r}}{{H}^{\circ }}=-y\text{ }kJ\text{ }mol{{e}^{-1}}$. And add equations (2) and (3) to get equation (1), then;
z= x-y
so , option(a) is incorrect.
(b) x=y-z
Consider the x equation:
$C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=x\text{ }kJ\text{ }mol{{e}^{-1}}$----------(4)
We have to make this equation from the rest two equations;
The rest two equation are as;
$CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$----------(5)
$C(graphite)+\dfrac{1}{2}{{O}_{2}}(g)\to CO(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=y\text{ }kJ\text{ }mol{{e}^{-1}}$----------(6)
On adding the equations (5) and (6), we will get the equation(4), then;
x=y +z
so, option(b) is also incorrect.
(c) x=y+ z
From the option (b), we come to know that the option (c) is correct.
(d) y=2z-x
Consider the y equation :
$C(graphite)+\dfrac{1}{2}{{O}_{2}}(g)\to CO(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=y\text{ }kJ\text{ }mol{{e}^{-1}}$---------(7)
We have to make this equation from the rest two equations;
The rest two equation are as;
$CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$----------(8)
$C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=x\text{ }kJ\text{ }mol{{e}^{-1}}$----------(9)
We can see that , in equation (7), CO is on the product side, we so will invert equation (8) so that the CO will come on product side and now it ${{\Delta }_{r}}{{H}^{\circ }}=-z\text{ }kJ\text{ }mol{{e}^{-1}}$. And add equations (8) and (9) to get equation (7), then;
y= x-z
so, option(d) is also incorrect.
Hence, from the above we can see that the correct option is (c).
Note: The enthalpy of formation of elements which are present in their molecular forms like oxygen gas, or in any solid form etc. their standard enthalpy of formation is always taken as zero as they undergo no change in their formation.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)