The reaction between hydrogen and chlorine gases gives hydrogen chloride gas in the following reaction:
H2(g) + Cl2(g) = 2HCl(g)
It is found experimentally that 1 mole of H2 gas reacts with 1 mole of Cl2 gas at constant temperature and pressure, to release 184 kJ of heat. What is the enthalpy change for this reaction?
∆h=2*(-184kJ mol^-1)
= -368kJ mol^-1
is this the right answer
I don't think so. dH rxn is -184 kJ.
thanks
Yes, that is the right answer. The enthalpy change (∆H) for a chemical reaction is the amount of heat energy released or absorbed during the reaction. In this case, it is given that 1 mole of H2 gas reacts with 1 mole of Cl2 gas, which means the stoichiometric coefficient for H2 in the balanced equation is 1, and for Cl2 it is also 1.
The given value of 184 kJ of heat released corresponds to 1 mole of H2 reacting. However, the balanced equation shows that 2 moles of HCl are formed for every mole of H2. Therefore, we need to multiply the given value by the stoichiometric coefficient of H2, which is 2.
So, the enthalpy change for the reaction is ∆H = 2 * (-184 kJ/mol) = -368 kJ/mol, which indicates that the reaction is exothermic (heat is being released).