Question

what is $delta_{r}h^{circ}$ for the following chemical reaction?
$h_{2}o(l)+ccl_{4}(l)
ightarrow cocl_{2}(g)+2hcl(g)$
you can use the following table of standard heats of formation ($delta_{f}h^{circ}$) to calculate the enthalpy of the given reaction.

element/ compound	standard heat of formation ($kj mol^{-1}$)	element/ compound	standard heat of formation ($kj mol^{-1}$)
$h_{2}(g)$	0	$o_{2}(g)$	0
$ccl_{4}(l)$	-139.5	$o(g)$	249
$h_{2}o(l)$	-285.8	$hcl(g)$	-92.30
$c(g)$	71	$cocl_{2}(g)$	-218.8
$c(s)$	0	$hno_{3}(aq)$	-206.6

express the standard enthalpy of reaction to three significant figures and include the appropriate units.

Explanation:

Step1: Recall the formula for $\Delta_{r}H^{\circ}$

$\Delta_{r}H^{\circ}=\sum n_{p}\Delta_{f}H^{\circ}(products)-\sum n_{r}\Delta_{f}H^{\circ}(reactants)$

Step2: Identify products and their $\Delta_{f}H^{\circ}$ values

For $COCl_{2}(g)$, $n_{1} = 1$ and $\Delta_{f}H^{\circ}(COCl_{2}(g))=- 218.8\ kJ/mol$. For $HCl(g)$, $n_{2}=2$ and $\Delta_{f}H^{\circ}(HCl(g))=-92.30\ kJ/mol$. So $\sum n_{p}\Delta_{f}H^{\circ}(products)=(-218.8)+2\times(-92.30)=-218.8 - 184.6=-403.4\ kJ/mol$

Step3: Identify reactants and their $\Delta_{f}H^{\circ}$ values

For $H_{2}O(l)$, $n_{3}=1$ and $\Delta_{f}H^{\circ}(H_{2}O(l))=-285.8\ kJ/mol$. For $CCl_{4}(l)$, $n_{4}=1$ and $\Delta_{f}H^{\circ}(CCl_{4}(l))=-139.5\ kJ/mol$. So $\sum n_{r}\Delta_{f}H^{\circ}(reactants)=(-285.8)+(-139.5)=-425.3\ kJ/mol$

Step4: Calculate $\Delta_{r}H^{\circ}$

$\Delta_{r}H^{\circ}=\sum n_{p}\Delta_{f}H^{\circ}(products)-\sum n_{r}\Delta_{f}H^{\circ}(reactants)=-403.4-(-425.3)=21.9\ kJ/mol$

Answer:

$21.9\ kJ/mol$