Calculate the standard enthalpy change for th | vedclass

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(C) 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

$-1355.00 \ kJ \ mol^{-1}$

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(C) 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 $C_2H_5OH_{(\ell)} + 3O_{2_{(g)}} ightarrow 2CO_{2_{(g)}} + 3H_2O_{(\ell)}$,the expression is:$\Delta_{r}H^{\circ} = [2 	imes \Delta_{f}H^{\circ}(CO_2) + 3 	imes \Delta_{f}H^{\circ}(H_2O)] - [\Delta_{f}H^{\circ}(C_2H_5OH) + 3 	imes \Delta_{f}H^{\circ}(O_2)]$.Since $\Delta_{f}H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state of an element),we have:$\Delta_{r}H^{\circ} = [2(-390) + 3(-285)] - [-280 + 3(0)]$.$\Delta_{r}H^{\circ} = [-780 - 855] - [-280]$.$\Delta_{r}H^{\circ} = -1635 + 280 = -1355 \ kJ \ mol^{-1}$.

Similar Questions

Calculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ Given that: $\Delta_{f} H^{\circ}(CH_4) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_2) = -394 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(H_2O) = -286 \ kJ \ mol^{-1}$

The correct order of $\Delta_{f} H^{\circ}$ values of Diamond $(I)$,Graphite $(II)$,and Fullerene $(III)$ is:

Which of the following reactions satisfies the following conditions?
$(a)$ Heat of combustion reaction
$(b)$ Heat of formation reaction
$(c)$ An exothermic reaction
$(d)$ Not a neutralization reaction

The heat of neutralization of a strong acid by a strong base is nearly equal to

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