What will be the amount of heat evolved by bu | vedclass

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

Question

(C) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.The enthalpy of combustion $\Delta H_c$ is calculated as

Vedclass · Accepted Answer

$322$

Vedclass · Answer

(C) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) ightarrow CO_2(g) + 2H_2O(l)$.The enthalpy of combustion $\Delta H_c$ is calculated as: $\Delta H_c = [\Delta H_f(CO_2) + 2 	imes \Delta H_f(H_2O)] - [\Delta H_f(CH_4) + 2 	imes \Delta H_f(O_2)]$.Given $\Delta H_f(O_2) = 0 \ kJ \ mol^{-1}$,we have:$\Delta H_c = -400 + 2(-240) - (-75) = -400 - 480 + 75 = -805 \ kJ \ mol^{-1}$.For $0.4 \ mol$ of $CH_4$,the heat evolved is: $0.4 \ mol 	imes 805 \ kJ \ mol^{-1} = 322 \ kJ$.

What will be the amount of heat evolved by burning $0.4 \ mol$ of methane? (Given heats of formation of $CH_4$,$CO_2$,and $H_2O$ are $-75$,$-400$,and $-240 \ kJ \ mol^{-1}$ respectively) ..... $kJ$

Similar Questions

The conversion of oxygen to ozone represented by the equation $3O_2 \to 2O_3$ is an endothermic reaction. The enthalpy change $\Delta H$ accompanying the reaction:

For the reaction,$3 C_2 H_{2(g)} \longrightarrow C_6 H_{6(g)}$,calculate the standard enthalpy change. The values of $\Delta H_f$ for $C_2 H_2$ and $C_6 H_6$ respectively are $250 \ kJ \ mol^{-1}$ and $90 \ kJ \ mol^{-1}$.

The enthalpy of neutralisation of $NH_4OH$ and $HCl$ is numerically:

Given that $\Delta H_f(H) = 218 \ kJ/mol$, express the $H-H$ bond energy in $kcal/mol$.

Enthalpy of formation of methane is $-75 \ kJ / mol$. What is the enthalpy change for formation of $24 \ g$ of methane (in $kJ$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module