Compound EX₄ reacts with Y₂ giving two products: EX₃Y and XY.Calculate the E-X bond enthalpy if the reaction liberates 243.6 kJ/mol and other bond enthalpies are: Y-Y 109 kJ/mol,X-Y 123.1 kJ/mol and E-Y 290 kJ/mol. A)60.5 kJ/mol B)547.7 kJ/mol C)20.6 kJ/mol D)234.7 kJ/mol E)176.9 kJ/mol

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The reaction involves breaking the E-X and Y-Y bonds and forming the E-Y and X-Y bonds. The enthalpy change for the reaction ($\Delta H$) is given as -243.6 kJ/mol (since it's exothermic, it liberates energy). The bond enthalpies are given for Y-Y (109 kJ/mol), X-Y (123.1 kJ/mol), and E-Y (290 kJ/mol). The bond enthalpy for E-X is unknown. The equation for the enthalpy change based on bond enthalpies is:$\Delta H = \text{Bonds broken} - \text{Bonds formed}$$\Delta H = (E-X + Y-Y) - (E-Y + X-Y)$Plugging in the known values:$-243.6 = (E-X + 109) - (290 + 123.1)$Solving for E-X gives:$E-X = -243.6 + 290 + 123.1 - 109 = 60.5 \text{ kJ/mol}$

Compound EX4 Reacts with Y2 Giving Two Products: EX3Y and XY.Calculate

Compound EX₄ Reacts with Y₂ Giving Two Products: EX₃Y and XY.Calculate