Benzene burns according to the following equations at 300K (R=8.314 J `"mole"^(-1)K^(-1))` `2C_(6)H_(6)(l) + 15O_(2)(g) rarr 12 CO_(2)(g) + 6H_(2)O(l)

Question

Benzene burns according to the following equations at 300K (R=8.314 J `"mole"^(-1)K^(-1))` `2C_(6)H_(6)(l) + 15O_(2)(g) rarr 12 CO_(2)(g) + 6H_(2)O(l)

Benzene burns according to the following equations at 300K (R=8.314 J `"mole"^(-1)K^(-1))`
`2C_(6)H_(6)(l) + 15O_(2)(g) rarr 12 CO_(2)(g) + 6H_(2)O(l) " " DeltaH^(@) =- 6542 KJ`
What is the `DeltaE^(@)` for the combustion of 1.5 mol of benzene
A. `-3271 KJ`
B. `-9812 KJ`
C. `-4906.5 `KJ
D. None of these

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
From given reaction `Deltan_(g) =12 -15=-3`
so `DeltaE^(@) =DeltaH^(@)-Deltan_(g)RT =-6542 + 3RT`
for 1.5 mole , `Delta E^(@) =(1.5)/(2) {-6542 + 3RT} = 4900 KJ`

4 views Answered 2023-02-05 15:29