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Question

The combustion of benzene  (l) gives $\ce{CO2(g)}$ and $\ce{H2O(l)}$. Given that the heat of combustion of benzene at constant volume is $\pu{-3263.9 kJ mol-1}$ at $\pu{25 ^\circ C}$; heat of combustion (in $\pu{kJ mol-1}$) of benzene at constant pressure will be: ($R = \pu{8.314 JK-1 mol-1}$)

Answer

$\Delta H = \pu{-3267.6 kJ}$

Explanation of Solution

$\ce{C6H6(l) + 15/2 O2 (g) -> 6CO2(g) + 3H2O(l)}$
$\Delta n_\mathrm g = 6- 7.5 = -1.5$
$\Delta U$ or $\Delta E = \pu{-3263.9 kJ}$
$\Delta H = \Delta U + \Delta n_\mathrm g RT$
So $\Delta H = -3263.9 + (-1.5) \times 8.314 \times 10^{-3} \times 298 = \pu{-3267.6 kJ}$

My question

When a reaction occurs at constant volume they equate enthalpy with internal energy which I think is wrong because according to the equation:

$$\mathrm dH = \mathrm dU +\mathrm d (PV)$$

At constant volume:

$$\mathrm dH = \mathrm dU + V\mathrm dP$$ In the question, heat of combustion is given = -3267.6 kj While solving for constant volume dU is given= -3267.6 kj

Why is $\mathrm dH = \mathrm dE$ at constant volume?

Question

The combustion of benzene(l) gives $\ce{CO2(g)}$ and $\ce{H2O(l)}.$ Given that the heat of combustion of benzene at constant volume is $\pu{-3263.9 kJ mol-1}$ at $\pu{25 ^\circ C}$; heat of combustion (in $\pu{kJ mol-1}$) of benzene at constant pressure will be: $(R = \pu{8.314 J K-1 mol-1})$

Answer

$\Delta H = \pu{-3267.6 kJ}$

Explanation of Solution

$\ce{C6H6(l) + 15/2 O2 (g) -> 6CO2(g) + 3H2O(l)}$
$\Delta n_\mathrm g = 6- 7.5 = -1.5$
$\Delta U$ or $\Delta E = \pu{-3263.9 kJ}$
$\Delta H = \Delta U + \Delta n_\mathrm g RT$
So $\Delta H = -3263.9 + (-1.5) \times 8.314 \times 10^{-3} \times 298 = \pu{-3267.6 kJ}$

My question

When a reaction occurs at constant volume they equate enthalpy with internal energy which I think is wrong because according to the equation:

$$\mathrm dH = \mathrm dU +\mathrm d(PV)$$

At constant volume:

$$\mathrm dH = \mathrm dU + V\mathrm dP$$ In the question, heat of combustion is given as $\pu{-3267.6 kJ},$ while solving for constant volume $\mathrm dU$ is given as $\pu{-3267.6 kJ}.$

Why is $\mathrm dH = \mathrm dE$ at constant volume?

Question

The combustion of benzene (l) gives $\ce{CO2(g)}$ and $\ce{H2O(l)}$. Given that the heat of combustion of benzene at constant volume is $\pu{-3263.9 kJ mol-1}$ at $\pu{25 ^\circ C}$; heat of combustion (in $\pu{kJ mol-1}$) of benzene at constant pressure will be: ($R = \pu{8.314 JK-1 mol-1}$)

Answer

$\Delta H = \pu{-3267.6 kJ}$

Explanation of Solution

$\ce{C6H6(l) + 15/2 O2 (g) -> 6CO2(g) + 3H2O(l)}$
$\Delta n_\mathrm g = 6- 7.5 = -1.5$
$\Delta U$ or $\Delta E = \pu{-3263.9 kJ}$
$\Delta H = \Delta U + \Delta n_\mathrm g RT$
So $\Delta H = -3263.9 + (-1.5) \times 8.314 \times 10^{-3} \times 298 = \pu{-3267.6 kJ}$

My question

When a reaction occurs at constant volume they equate enthalpy with internal energy which I think is wrong because according to the equation:

$$\mathrm dH = \mathrm dU +\mathrm d (PV)$$

At constant volume:

$$\mathrm dH = \mathrm dU + V\mathrm dP$$

Why is $\mathrm dH = \mathrm dE$ at constant volume?

Question

The combustion of benzene (l) gives $\ce{CO2(g)}$ and $\ce{H2O(l)}$. Given that the heat of combustion of benzene at constant volume is $\pu{-3263.9 kJ mol-1}$ at $\pu{25 ^\circ C}$; heat of combustion (in $\pu{kJ mol-1}$) of benzene at constant pressure will be: ($R = \pu{8.314 JK-1 mol-1}$)

Answer

$\Delta H = \pu{-3267.6 kJ}$

Explanation of Solution

$\ce{C6H6(l) + 15/2 O2 (g) -> 6CO2(g) + 3H2O(l)}$
$\Delta n_\mathrm g = 6- 7.5 = -1.5$
$\Delta U$ or $\Delta E = \pu{-3263.9 kJ}$
$\Delta H = \Delta U + \Delta n_\mathrm g RT$
So $\Delta H = -3263.9 + (-1.5) \times 8.314 \times 10^{-3} \times 298 = \pu{-3267.6 kJ}$

My question

When a reaction occurs at constant volume they equate enthalpy with internal energy which I think is wrong because according to the equation:

$$\mathrm dH = \mathrm dU +\mathrm d (PV)$$

At constant volume:

$$\mathrm dH = \mathrm dU + V\mathrm dP$$ In the question, heat of combustion is given = -3267.6 kj While solving for constant volume dU is given= -3267.6 kj

Why is $\mathrm dH = \mathrm dE$ at constant volume?

I'm very confused regarding this... When a reaction occurs at constant volume they equate enthalpy with internal energy which I think is wrong because according to the equation:-

$dH = dU +d (PV)$

At constant volume:-

$dH = dU + V*dP$

see in question no. 7 It is given:- dH=dE please explain this...

Question

The combustion of benzene (l) gives $\ce{CO2(g)}$ and $\ce{H2O(l)}$. Given that the heat of combustion of benzene at constant volume is $\pu{-3263.9 kJ mol-1}$ at $\pu{25 ^\circ C}$; heat of combustion (in $\pu{kJ mol-1}$) of benzene at constant pressure will be: ($R = \pu{8.314 JK-1 mol-1}$)

Answer

$\Delta H = \pu{-3267.6 kJ}$

Explanation of Solution

$\ce{C6H6(l) + 15/2 O2 (g) -> 6CO2(g) + 3H2O(l)}$
$\Delta n_\mathrm g = 6- 7.5 = -1.5$
$\Delta U$ or $\Delta E = \pu{-3263.9 kJ}$
$\Delta H = \Delta U + \Delta n_\mathrm g RT$
So $\Delta H = -3263.9 + (-1.5) \times 8.314 \times 10^{-3} \times 298 = \pu{-3267.6 kJ}$

My question

When a reaction occurs at constant volume they equate enthalpy with internal energy which I think is wrong because according to the equation:

$$\mathrm dH = \mathrm dU +\mathrm d (PV)$$

At constant volume:

$$\mathrm dH = \mathrm dU + V\mathrm dP$$

Why is $\mathrm dH = \mathrm dE$ at constant volume?