The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is −3263.9 kJ mol−1 at 25∘C; heat of combustion (in kJmol−1 ) of benzene at constant pressure will be (R=8.314JK−1 mol−1)

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The combustion of benzene (l) gives {CO}_{2}(g) and {H}_{2} {O

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The combustion of benzene $(l)$ gives $CO_{2} (g)$ and $H_{2} O (l)$ . Given that heat of combustion of benzene at constant volume is $- 3263.9 kJ mol^{- 1}$ at $2 5^{\circ} C$ ; heat of combustion (in $kJ mol^{- 1}$ ) of benzene at constant pressure will be $(R = 8.314 JK^{- 1} mol^{- 1})$

4152.6

- 452.46

3260

- 3267.6

Views: 6,237 students

Updated on: Feb 23, 2023

Book:

Arihant Chemistry JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

Found 4 tutors discussing this question

Alexander Discussed

The combustion of benzene

(l)

gives

CO_{2} (g)

and

H_{2} O (l)

. Given that heat of combustion of benzene at constant volume is

- 3263.9 kJ mol^{- 1}

2 5^{\circ} C

; heat of combustion (in

kJ mol^{- 1}

) of benzene at constant pressure will be

(R = 8.314 JK^{- 1} mol^{- 1})

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

Exp . (d)

Key Idea Calculate the heat of combustion with the help of following formula

Δ H_{p} = Δ U + Δ n_{g} RT

where,

Δ H_{p} =

Heat of combustion at constant pressure

Δ U =

Heat at constant volume (It is also called

Δ E

Δ n_{g} =

Change in number of moles (In gaseous state).

R =

Gas constant;

T =

Temperature. From the equation,

C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) ⟶ 6 CO_{2} (g) + 3 H_{2} O (l)

Change in the number of gaseous moles i.e.

Δ n_{g} = 6 - \frac{15}{2} = - \frac{3}{2} or - 1.5

Now we have

Δ n_{g}

and other values given in the question are

Δ U = - 3263.9 kJ / mol

T = 2 5^{\circ} C = 273 + 25 = 298 K

R = 8.314 JK^{- 1} mol^{- 1}

So, Δ H_{p} = (- 3263.9) + (- 1.5) \times 8.314

\times 1 0^{- 3} \times 298

= - 3267.6 kJ mol^{- 1}

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Question Text	The combustion of benzene $(l)$ gives $CO_{2} (g)$ and $H_{2} O (l)$ . Given that heat of combustion of benzene at constant volume is $- 3263.9 kJ mol^{- 1}$ at $2 5^{\circ} C$ ; heat of combustion (in $kJ mol^{- 1}$ ) of benzene at constant pressure will be $(R = 8.314 JK^{- 1} mol^{- 1})$
Updated On	Feb 23, 2023
Topic	Thermodynamics and Thermochemistry
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 12
Upvotes	1120
Avg. Video Duration	7 min

The combustion of benzene $(l)$ gives $CO_{2} (g)$ and $H_{2} O (l)$ . Given that heat of combustion of benzene at constant volume is $- 3263.9 kJ mol^{- 1}$ at $2 5^{\circ} C$ ; heat of combustion (in $kJ mol^{- 1}$ ) of benzene at constant pressure will be $(R = 8.314 JK^{- 1} mol^{- 1})$

Text solutionVerified

Filo tutor solutions (12)

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The combustion of benzene (l) gives CO2​(g) and H2​O(l). Given that heat of combustion of benzene at constant volume is −3263.9 kJ mol−1 at 25∘C; heat of combustion (in kJmol−1 ) of benzene at constant pressure will be (R=8.314JK−1 mol−1)

Text solutionVerified

Filo tutor solutions (12)

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The combustion of benzene $(l)$ gives $CO_{2} (g)$ and $H_{2} O (l)$ . Given that heat of combustion of benzene at constant volume is $- 3263.9 kJ mol^{- 1}$ at $2 5^{\circ} C$ ; heat of combustion (in $kJ mol^{- 1}$ ) of benzene at constant pressure will be $(R = 8.314 JK^{- 1} mol^{- 1})$