7.Calculate the volume of nitrogen that reacts with 12dm3 of hydrogen with the volume of both gases measured at rtp: 3H2 + N2 > 2NH3

4.03dm³

Explanation:

The reaction expression is given as:

       3H₂   +   N₂   →    2NH₃  

  Volume of hydrogen  = 12dm³  

AT rtp:

             1 mole of gas occupies volume of 22.4dm³  

             x mole of hydrogen will occupy a volume of 12dm³

     Number of moles of hydrogen  =   = 0.54mole

From the balanced reaction equation:

            3 mole of hydrogen gas combines with 1 mole of Nitrogen gas

         0.54 mole of hydrogen as will therefore combine with   = 0.18moles of nitrogen gas

Since ;

                     1 mole of gas occupies a volume of 22.4dm³

               0.18moles of Nitrogen gas will occupy 0.18 x 22.4  = 4.03dm³

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.