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1

There's a general principle that applies anytime you calculate a change in a property: The change in the value of a property in going from an initial state to a final state is always value(final state) - value(initial state). For instance, what is the change in elevation (which we'll call height, h; not to be confused with enthalpy, H!) when we go from 500 ...


0

The enthalpy change is always Products minus Reactants.


1

Maurice has stated why equation 1 is much more exothermic than equation 2. A visualization of the processes may help you understand the details. Equation 1 involves the reaction of a carbon atom with four atoms of hydrogen to form methane in the gas phase. The bond dissociation energy (BDE) of the hydrogen molecule is +104 kcal/mol. Accordingly, the heat of ...


0

The heat of formation of $CH_4$ is $-75$ kJ/mol. This is not much : y is small. As the dissociation energies of $C-H$ bonds are higher than $400 kJ/mol$, the heat of reaction of the first equation is more than $1600 kJ/mol$. This first reaction is about $20$ times more exothermic than the second reaction. x >> y !!


0

To understand internal energy*, I prefer to start with this: Consider a thermodynamic system. There are only three ways we can increase its internal energy (which is the sum of its internal potential energy and internal kinetic energy): flow heat into it; do work on it; or add matter to it (and the converse for decreasing its internal energy). If the ...


0

The internal energy is defined in (classical) thermodynamics for a finite change as $\Delta U=q+w$ where $U$ is the energy contained in the 'system' called the internal energy, and $q$ is the heat absorbed by the system and $w$ the work done on the system. Often the word 'system' means some ideal gas (often imagined to be in a cylinder with a frictionless ...


-2

The internal energy U is the sum of all energies stored in the chemical bonds of the examined sample of matter. It is a potential energy. The enthalpy H is the same, but with a correction due to the presence of the atmosphere. The difference between H and U is similar to the difference between weight and apparent weight. Let me go back to this measurement. ...


1

$$\ce{4 HCOOH + O2 -> 4 CO2 + 2 H2O}$$ Your reaction equation is wrong. Just count the $\ce{H}$'s on both sides. The actual equation is $$\ce{2 CH2O2 + O2 -> 2 CO2 + 2 H2O}$$ which results in the correct enthalpy of combustion. $$\begin{align}-2\times-393.5\ \mathrm{kJ/mol}-2\times-285.8\ \mathrm{kJ/mol} +2\times-425.0\ \mathrm{kJ/mol}&=508.6\ ...


1

Option D is the only one corresponding to the definition "energy required to break one mole of a specific bond" and nothing more.


1

In order to make the phase transition, you need to supply your sample molecules with an additional amount of kinetic energy (i.e., the heat of fusion or vaporization) to overcome their intermolecular bonding. This however means that after the transition, your system contains more internal energy than before. That additional energy is encoded in the vertical ...


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