At constant pressure and 25 $\circ$ C,what is $\Delta$_rH $\circ$ for the following reaction 2C₂H₆(g)+ 7O₂(g)$\to$ 4CO₂(g)+ H₂O(l) If the complete consumption of 89.4 g of C₂H₆ liberates -4638 kJ of heat energy?

Question

Quizplus · Accepted Answer

The enthalpy change for the reaction can be calculated using the formula:$$\Delta H = -q$$where q is the heat energy liberated.In this case, q = -4638 kJ and the number of moles of C₂H₆ consumed is:$$89.4 g C_2H_6 imes \frac{1 mol C_2H_6}{30.07 g C_2H_6} = 2.97 mol C_2H_6$$Therefore, the enthalpy change per mole of reaction is:$$\Delta H_{rxn} = \frac{-4638 kJ}{2.97 mol} = -1560 kJ/mol-rxn$$The correct choice is A.

At Constant Pressure and 25 ∘\circ∘ C,what Is Δ\DeltaΔ rH ∘\circ∘ for the Following Reaction 2C2H6(g)+ 7O2(g)

At Constant Pressure and 25^$\circ$C,what Is $\Delta$ _rH^$\circ$ for the Following Reaction
2C₂H₆(g)+ 7O₂(g)