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# Given that:(i) $C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=x\text{ }kJ\text{ }mol{{e}^{-1}}$ (ii) $C(graphite)+\dfrac{1}{2}{{O}_{2}}(g)\to CO(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=y\text{ }kJ\text{ }mol{{e}^{-1}}$(iii) $CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g)\text{ }{{\Delta }_{r}}{{H}^{\circ }}=z\text{ }kJ\text{ }mol{{\text{e}}^{-1}}$Based on the given thermodynamic equations, find out which one of the following algebraic relationships is correct?(a) z= x +y(b) x=y-z(c) x=y+ z(d) y=2z- x

Last updated date: 20th Jun 2024
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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.

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.