The heats of combustion of yellow phosphorus | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The combustion reactions are as follows:$1$) $P_{	ext{yellow}} + O_2 ightarrow P_4O_{10} \quad \Delta H_1 = -9.91 \, kJ$$2$) $P_{	ext{red}} +

Vedclass · Accepted Answer

$-1.13$

Vedclass · Answer

(D) The combustion reactions are as follows:$1$) $P_{	ext{yellow}} + O_2 ightarrow P_4O_{10} \quad \Delta H_1 = -9.91 \, kJ$$2$) $P_{	ext{red}} + O_2 ightarrow P_4O_{10} \quad \Delta H_2 = -8.78 \, kJ$We want the transition: $P_{	ext{yellow}} ightarrow P_{	ext{red}}$This can be obtained by subtracting equation $(2)$ from equation $(1)$:$(P_{	ext{yellow}} + O_2) - (P_{	ext{red}} + O_2) = -9.91 - (-8.78)$$P_{	ext{yellow}} - P_{	ext{red}} = -9.91 + 8.78$$P_{	ext{yellow}} ightarrow P_{	ext{red}} \quad \Delta H_{	ext{transition}} = -1.13 \, kJ$

The heats of combustion of yellow phosphorus and red phosphorus are $-9.91 \, kJ$ and $-8.78 \, kJ$ respectively. The heat of transition for the conversion of yellow phosphorus to red phosphorus will be ...... $kJ$?

Similar Questions

The heat of combustion of $C$,$S$ and $CS_2$ are $-393.3 \ kJ$,$-293.7 \ kJ$ and $-1108.76 \ kJ$ respectively. What will be the heat of formation of $CS_2$ in $kJ$?

Enthalpy of a compound is equal to its

Explain the standard enthalpy of formation.

If the bond formation energy of the $H-H$ bond is $-433 \ kJ \ mol^{-1}$,find the bond dissociation energy for $0.5 \ mol$ of $H_{2(g)}$. (in $kJ$)

Identify the $INVALID$ equation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module