Question

ons: $delta h_1=-393.5\text{kj}$ $delta h_2 = - 566.0\text{kj}$ $delta h_3 = 483.6\text{kj}$ $delta h_{rxn}=?$ what is the overall enthalpy of reaction? $delta h_{rxn}=square\text{kj}$ the second equation must be reversed. the first equation must be halved.

Explanation:

Step1: Adjust $\Delta H_1$

Since the first equation must be halved, the new $\Delta H$ for the first - adjusted equation is $\Delta H_{1_{new}}=\frac{\Delta H_1}{2}=\frac{- 393.5}{2}=-196.75$ kJ.

Step2: Adjust $\Delta H_2$

Since the second equation must be reversed, the new $\Delta H$ for the second - adjusted equation is $\Delta H_{2_{new}}=- \Delta H_2=-(-566.0) = 566.0$ kJ.

Step3: $\Delta H_3$ remains the same

$\Delta H_{3_{new}}=\Delta H_3 = 483.6$ kJ.

Step4: Calculate $\Delta H_{rxn}$

$\Delta H_{rxn}=\Delta H_{1_{new}}+\Delta H_{2_{new}}+\Delta H_{3_{new}}=-196.75 + 566.0+483.6$
$=-196.75+1049.6 = 852.85$ kJ

Answer:

$852.85$