At 25°C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are –3268 kJ mol–1 and –1300 kJ mol–1, respectively. The change in enthalpy for the reaction 3C2H2(g) → C6H6(I), is
- +324 kJ mol–1
- +632 kJ mol–1
- –632 kJ mol–1
- –732 kJ mol–1
The Correct Option is A
Solution and Explanation
The correct option is(C): –632 kJ mol–1.
I.
\(C_6H_6(l)+\frac{15}{2}O_2(g)→6CO_2(g)+3H_2O(g)\)
ΔH1 = –3268 kJ/mol
II.
\(C_2H_2(g)+\frac{5}{2}O_2(g)→2CO_2(g)+H_2O(g)\)
ΔH2 = –1300 kJ/mol
III.
\(3C_2H_2(g)-2C→6H_6(l)\)
ΔH3
Applying Hess’s law of constant heat summation
ΔH3 = 3 × ΔH2 – ΔH1
= 3 × (–1300) – (–3268)
= –632 kJ/mol
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Concepts Used:
Enthalpy change
Enthalpy Change refers to the difference between the heat content of the initial and final state of the reaction. Change in enthalpy can prove to be of great importance to find whether the reaction is exothermic or endothermic.
Formula for change in enthalpy is:-
dH = dU + d(PV)
The above equation can be written in the terms of initial and final states of the system which is defined below:
UF – UI = qP –p(VF – VI)
Or qP = (UF + pVF) – (UI + pVI)
Enthalpy (H) can be written as H= U + PV. Putting the value in the above equation, we obtained:
qP = HF – HI = ∆H
Hence, change in enthalpy ∆H = qP, referred to as the heat consumed at a constant pressure by the system. At constant pressure, we can also write,
∆H = ∆U + p∆V
Standard Enthalpy of Reaction
To specify the standard enthalpy of any reaction, it is calculated when all the components participating in the reaction i.e., the reactants and the products are in their standard form. Therefore the standard enthalpy of reaction is the enthalpy change that occurs in a system when a matter is transformed by a chemical reaction under standard conditions.