Calculate the standard enthalpy change of the | vedclass

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

Question

(A) The standard enthalpy change of the reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(	ext{products}) - \

Vedclass · Accepted Answer

$-1411.1 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The standard enthalpy change of the reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(	ext{products}) - \sum \Delta_{f}H^{\circ}(	ext{reactants})$ For the reaction: $C_2H_{4(g)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 2H_2O_{(\ell)}$ $\Delta_{r}H^{\circ} = [2 	imes \Delta_{f}H^{\circ}(CO_2) + 2 	imes \Delta_{f}H^{\circ}(H_2O)] - [1 	imes \Delta_{f}H^{\circ}(C_2H_4) + 3 	imes \Delta_{f}H^{\circ}(O_2)]$ Since $\Delta_{f}H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state): $\Delta_{r}H^{\circ} = [2(-393.5) + 2(-285.8)] - [52 + 0]$ $\Delta_{r}H^{\circ} = [-787 - 571.6] - 52$ $\Delta_{r}H^{\circ} = -1358.6 - 52 = -1410.6 \ kJ \ mol^{-1}$ The closest value is $-1411.1 \ kJ \ mol^{-1}$.

Similar Questions

For which of the following reactions is the $\Delta H^o$ of the reaction equal to the $\Delta H_f^o$ of the product?

From the given diagram, the enthalpy change $\Delta_r H$ for the reaction $C \rightarrow A$ is:

The enthalpy of solution of $NaOH_{(s)}$ in water is $-41.6 \ kJ \ mol^{-1}$. When $NaOH_{(s)}$ is dissolved in water,the temperature of the water will ..........

The heat of complete neutralization of $1 \ mol$ of $H_2SO_4$ with a strong base is ....... $Kcal$.

What is the standard $N \equiv N$ bond enthalpy from the following reaction? $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta H^{\circ} = -83 \ kJ$
Given: $(\Delta H^{\circ}_{(H-H)} = 435 \ kJ \text{ mol}^{-1}, \Delta H^{\circ}_{(N-H)} = 389 \ kJ \text{ mol}^{-1})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module