Question

Calculate the enthalpy change for the reaction: 4Al(s) + 3MnO2(s) --> 2Al2O3 (s) + 3Mn(s)2Al(s) + 3/2 O2 (g) --> Al2O3(s). ∆ H=-1676.0 kJMn(s) + O2 (g) --> MnO2(s). ∆ H = -521.0 kJ

Answer 1

Step-by-step explanation:

To solve this question, we need to use the Hess's Law.

Hess's Law states that: in a chemical reaction, the heat released or absorbed is constant and independent of the number of steps the reaction goes through. That is, the law states that the enthalpy change of a chemical reaction depends only on its initial state and its final state.

Hess's Law is also known as the law of the sum of heats of reaction, because the enthalpy change is equal to the sum of the changes in the steps through which the chemical reaction passes (intermediate reactions). The calculation is performed as follows:

- If the chemical reaction is inverted, the sign of the enthalpy change must also be inverted;

- If the equation is multiplied, the enthalpy change must also be multiplied;

- If dividing the equation, the enthalpy change must also be divided.

So we're going to invert, divide, or multiply each given equation in order to get the equation that the question gives us.

The equation the question gives us is:

4Al(s) + 3MnO2(s) --> 2Al2O3 (s) + 3Mn(s)

Let's multiply the first equation by 2:

4 Al + 3 O2 --> 2 Al2O3

ΔH = -1676 * 2 = -3,352 kJ

Let's invert and multiply by 3 the second equation:

3 MnO2 -> 3 Mn + 3O2

ΔH = -521 * 3 * (-1)

ΔH = 1,563 kJ

Answer: -1,789 kJ