Question

X2 + 3Y2 --> 2XY3 triangle H1= -320 kJ
X2 + 2Z2 --> 2xZ2 triangle H2= -170 kJ
2Y2 + Z2 --> 2Y2Z triangle H3= -250 kJ
Calculate the change in enthalpy for the following reaction:
4XY3 + 7Z2 --> 6Y2Z + 4XZ2
Triangle H =____?___ kJ

Answer 1

Answer on Question #41036 - Chemistry - Other

Question:

$$\begin{array}{l} X_2 + 3Y_2 \rightarrow 2XY_3 \quad \Delta H1 = -320\ \mathrm{kJ} \\ X_2 + 2Z_2 \rightarrow 2XZ_2 \quad \Delta H2 = -170\ \mathrm{kJ} \\ 2Y_2 + Z_2 \rightarrow 2Y_2Z \quad \Delta H3 = -250\ \mathrm{kJ} \\ \end{array}$$

Calculate the change in enthalpy for the following reaction:

$$4XY_3 + 7Z_2 \rightarrow 6Y_2Z + 4XZ_2$$ $$\Delta H = \_ \_ \_ ? \_ \_ \mathrm{kJ}$$

Solution.

$$\begin{array}{l} X_2 + 3Y_2 \rightarrow 2XY_3 \quad (1) \\ X_2 + 2Z_2 \rightarrow 2XZ_2 \quad (2) \\ 2Y_2 + Z_2 \rightarrow 2Y_2Z \quad (3) \\ 4XY_3 + 7Z_2 \rightarrow 6Y_2Z + 4XZ_2 \quad (4) \\ \end{array}$$

It can easily be seen, that the fourth equation can be expressed through the sum of first three in such way:

$$(4) = -2*(1) + 2*(2) + 3*(3)$$

According to Hess's law, the change in enthalpy for the reaction is:

$$\Delta H = 3 * \Delta H_3 + 2 * \Delta H_2 - 2 * \Delta H_1 = -250 * 3 - 170 * 2 + 320 * 2 = -450\ \mathrm{kJ}$$

Answer: $\Delta H = -450\ \mathrm{kJ}$