Enthalpy calculation

Question

What is the value of enthalpy change for the following reaction

$$\ce{X + 2Y -> 2Z}$$

given

\begin{align} \ce{W + X &-> 2Y} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-200 kcal} \tag{1}\\ \ce{2W + 3X &-> 2Z + 2Y} &\qquad & \Delta_\mathrm{r}H_2 = \pu{-150 kcal} \tag{2} \end{align}

I have the answers, and the correct answer is $\pu{250 kcal}$, so I'm assuming you do $400-150$, but not sure how you would go about it.

Answer 1

$$\ce{X + 2Y -> 2Z}$$

\begin{align} \ce{W + X &-> 2Y} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-200 kcal} \tag{1}\\ \ce{2W + 3X &-> 2Z + 2Y} &\qquad & \Delta_\mathrm{r}H_2 = \pu{-150 kcal} \tag{2} \end{align}

$(-1) \times 2$ + $(2)$ gives:

$\ce{4Y +2W + 3X->2W + 2X + 2Z + 2Y} \equiv \ce{2Y +X -> 3Z}$

Using Hess's Law, the enthalpy change for the required reaction is given by $-2\times \Delta_rH_1+\Delta_rH_2= 400-150 = \pu{250 kcal}$

Answer 2

Keep in mind you are allowed to perform basic math operations (+,–,×,÷) with the balanced equations of chemical reactions. I'd recommend to start with undesired component(s) ($\ce{W}$) which should not appear in final equation, and try to cancel them out.

Here, to obtain target reaction and eliminate $\ce{W}$, you need to subtract doubled first reaction from the second one (e.g. "$(2) - 2 \cdot (1)$"), which results in

\begin{align} \require{cancel} \ce{\cancel{\ce{2W}} + 3X - \cancel{\ce{2W}} - 2X &-> 2Z + 2Y - 4Y} &\qquad &\Delta_\mathrm{r}H = \Delta_\mathrm{r}H_2 - 2\Delta_\mathrm{r}H_1 \\ \ce{X + 2Y &-> 2Z} &\qquad &\Delta_\mathrm{r}H = \Delta_\mathrm{r}H_2 - 2\Delta_\mathrm{r}H_1 \end{align}

$$\Delta_\mathrm{r}H = \Delta_\mathrm{r}H_2 - 2\Delta_\mathrm{r}H_1 = \pu{-150 kcal} -2 \cdot (\pu{-200 kcal}) = \pu{250 kcal}$$