The reaction between potassium metal and chlorine gas is described by the following thermochemical equation K(s) + ½C_l2(g) = KCl(s) ÄH =–437 kJ mol–1

What will be the enthalpy change (to 2 significant figures) for the reaction when 6.0 g of potassium metal reacts with chlorine gas at constant pressure? The relative atomic mass of potassium is 39. ÄH =

6 g is what part of a mole if a mole of K is 39 grams?

delta H = -437 kJ/mol for the reaction. So it will be how many kJ for the part of the mole K you have?

To calculate the enthalpy change for the reaction, we first need to convert the given mass of potassium into moles. Here are the steps to follow:

Step 1: Find the molar mass of potassium (K)
- The relative atomic mass of potassium is given as 39 g/mol.

Step 2: Calculate the number of moles of potassium (n)
- Use the formula: n = mass/molar mass
- Plug in the values: mass = 6.0 g and molar mass = 39 g/mol
- Calculate: n = 6.0 g / 39 g/mol

Step 3: Determine the number of moles of chlorine gas (Cl₂)
- Since the stoichiometric coefficient of chlorine gas is ½ in the balanced equation, we need to find the number of moles of Cl₂.
- Multiply the number of moles of potassium by the coefficient ratio.
- In this case, the coefficient ratio is 1:1 since the coefficient for potassium (K) is 1 and for Cl₂ is ½.
- Therefore, the moles of chlorine gas will be the same as moles of potassium.

Step 4: Calculate the enthalpy change
- The enthalpy change (ΔH) is given as -437 kJ/mol in the thermochemical equation.
- Multiply the number of moles of the reactant by the enthalpy change to get the total enthalpy change.
- In this case, the enthalpy change will be the same as ΔH since the mole ratio is 1:1.
- Therefore, the enthalpy change will be: ΔH = -437 kJ/mol * (number of moles of Cl₂ or K)

Let's calculate the enthalpy change using these steps:

Step 1: Molar mass of potassium (K) = 39 g/mol

Step 2: Number of moles of potassium (K) = 6.0 g / 39 g/mol = 0.154 mol

Step 3: Number of moles of chlorine gas (Cl₂) = 0.154 mol

Step 4: Enthalpy change (ΔH) = -437 kJ/mol * 0.154 mol
= -67.198 kJ

Therefore, the enthalpy change for the reaction between 6.0 g of potassium metal and chlorine gas at constant pressure will be -67.2 kJ (rounded to 2 significant figures).