For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, `Delta_CH^(theta)=-601.70kJmol^(-1)`, the magnitude of change in internal energy for the reaction is __________ kJ . (Nearest integer)
(Given : `R= 8.3 J K^(-1) mol^(-1)`)

Explanation

To find the magnitude of the change in internal energy (ΔU) for the combustion of one mole of magnesium, we can use the relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) given by the equation:

\[ \Delta H = \Delta U + \Delta n_g RT \]

Where:
- \(\Delta H\) is the change in enthalpy (given as -601.70 kJ/mol).
- \(\Delta U\) is the change in internal energy (which we need to find).
- \(\Delta n_g\) is the change in the number of moles of gas during the reaction.
- \(R\) is the universal gas constant (given as 8.3 J K\(^{-1}\) mol\(^{-1}\)).
- \(T\) is the temperature in Kelvin (given as 300 K).

### Step 1: Determine the change in the number of moles of gas (\(\Delta n_g\))

The balanced equation for the combustion of magnesium is:

\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

In this reaction:
- Reactants: 1 mole of \(\text{O}_2\) (gas).
- Products: 0 moles of gas (since \(\text{MgO}\) is a solid).

Thus, the change in the number of moles of gas is:

\[ \Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 0 - 1 = -1 \]

### Step 2: Substitute the values into the equation

Now we can substitute the values into the equation:

\[ \Delta H = \Delta U + \Delta n_g RT \]

Substituting the known values:

\[ -601.70 \text{ kJ} = \Delta U + (-1) \times (8.3 \text{ J K}^{-1} \text{ mol}^{-1}) \times (300 \text{ K}) \]

### Step 3: Calculate \(\Delta n_g RT\)

Calculating \(\Delta n_g RT\):

\[ \Delta n_g RT = -1 \times 8.3 \text{ J K}^{-1} \text{ mol}^{-1} \times 300 \text{ K} = -2490 \text{ J} = -2.49 \text{ kJ} \]

### Step 4: Rearranging the equation to find \(\Delta U\)

Now we can rearrange the equation to find \(\Delta U\):

\[ \Delta U = -601.70 \text{ kJ} - (-2.49 \text{ kJ}) \]
\[ \Delta U = -601.70 \text{ kJ} + 2.49 \text{ kJ} \]
\[ \Delta U = -599.21 \text{ kJ} \]

### Step 5: Find the magnitude of \(\Delta U\)

Since we are asked for the magnitude, we take the absolute value:

\[ |\Delta U| = 599.21 \text{ kJ} \]

Rounding to the nearest integer gives us:

\[ |\Delta U| \approx 599 \text{ kJ} \]

### Final Answer

The magnitude of change in internal energy for the reaction is **599 kJ**.

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