During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is

(i)

2C_{4}H_{10}(g)+13O_{2}(g)\rightarrow 8CO_{2}(g)+10H_{2}O_{(l)}\Delta _{c}H=-2658.0\; kJ\; mol^{-1}

(ii)

C_{4}H_{10}(g)+\frac{13}{2}O_{2}(g)\rightarrow 4CO_{2}(g)+5H_{2}O(g)\Delta _{c}H=-1329.0\; kJ\; mol^{-1}

(iii)

C_{4}H_{10}(g)+\frac{13}{2}O_{2}(g)\rightarrow 4CO_{2}(g)+5H_{2}O(l)\Delta _{c}H=-2658.0\; kJ\; mol^{-1}

(iv)

C_{4}H_{10}(g)+\frac{13}{2}O_{2}(g)\rightarrow 4CO_{2}(g)+5H_{2}O(g)\Delta _{c}H=+2658.0 kJ\; mol^{-1}

Answers (1)

The answer is the option (iii).

Explanation: The definition of the enthalpy of combustion is when the substance undergoes combustion and the enthalpy change that happens per mole in that substance. All other reactants are assumed to be in their standard states at that time.

Most Viewed Questions

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions