The enthalpy of the formation of propane can be calculated using Hess’s Law, which states that the enthalpy change of a chemical reaction is independent of the route by which the chemical reaction takes place. This is done through the manipulation of the enthalpy values provided. For propane, we know the balanced equation for its formation is: C(s) + H₂(g) \(\rightarrow\) C₃H₈(l)
But we'll need 3 moles of carbon and 4 moles of hydrogen for this, so the equation should be: 3C(s) + 4H₂(g) \(\rightarrow\) C₃H₈(l)
Hence, according to Hess’s Law, the enthalpy change of the formation of propane is:
ΔH = [ 1 x (–2220.0 kJ mol⁻¹) - 3 x (–393.5 kJ mol⁻¹) - 4 x (–285.8 kJ mol⁻¹)]
= –103.5 kJ mol⁻¹
\(\Rightarrow\) round off to the nearest whole number
\(\Rightarrow\) -104 kJ mol⁻¹
So, the correct answer is 104
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is:
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
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.
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
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.