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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.

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  • 1
    $\begingroup$ FWIW, a direct implication of Hess's Law is that you can superpose those equations, which allows to treat those equations as linear equations of chemical energy; mathematically, you can said that $w + x - 2y = - 200$, $2w+3x - 2y - 2z = -150$, which gives you a nice $\begin{matrix}1 & 1 & -2 & 0 & -200\\2 & 3 & -2 & -2 & -150\end{matrix} $ matrix, which you'd need to transform to get $\begin{matrix}0 & 1 & 2 & -2 & result\end{matrix}$ form - if it's impossible (the usual caveats from en.wikipedia.org/wiki/System_of_linear_equations apply), then the set of equations is unsolvable. $\endgroup$ – vaxquis Nov 26 '17 at 13:00
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$$\ce{X + 2Y -> 2Z}$$

\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 $-2\times \Delta_rH_1+\Delta_rH_2= 400-150 = \pu{250 kcal}$

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