H2(g)+Cl2(g)=2HCL(g)
it is found experimentally that 1 mole of H2 gas react with 1 mole of CL2 gas at constant temperature & pressure , to release 184kJ of heat .what is the enthalpy change for this reaction?
∆H=
-184 kJ/reaction.
+184 kj
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To calculate the enthalpy change (∆H) for the given reaction, we need to use the equation:
∆H = ∑∆H(products) - ∑∆H(reactants)
Where ∆H(products) is the sum of the enthalpies of the products and ∆H(reactants) is the sum of the enthalpies of the reactants.
In this reaction, we have:
Reactants:
H2 (g) with 1 mole of H2 gas
Cl2 (g) with 1 mole of Cl2 gas
Products:
2HCl (g) with 2 moles of HCl gas
Given that 1 mole of H2 gas reacts with 1 mole of Cl2 gas to release 184 kJ of heat, we can determine that the enthalpy change for the reaction is -184 kJ/mol (negative because heat is released).
Therefore, ∆H = -184 kJ/mol