What is the ∆H value for this thermo chemical equation? Full question in the description box below.

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2 Answers
Jan 10, 2018

Answer:

#sf(-741color(white)(x)kJ"/""mol")#

Explanation:

#sf(M_r[C_3H_8]=44.1)#

We are told that:

1 tonne gives #sf(1.68xx10^4color(white)(x)MJ)#

#:.##sf(cancel(10^6)color(white)(x)grarr1.68xx10^4xxcancel(10^(6))J)#

This means that:

#sf(1color(white)(x)grarr1.68xx10^4color(white)(x)J)#

We need to find the energy released by 1 mole which is 44.1 g:

#:.##sf(44.1color(white)(x)grarr1.68xx10^4xx44.1color(white)(x)J)#

#sf(=74.088xx10^4color(white)(x)J)#

#sf(=741color(white)(x)"kJ")#

Since heat is released we can say that:

#sf(DeltaH=-741color(white)(x)"kJ""/""mol")#

Jan 10, 2018

Answer:

#DeltaH=-739# #kJmol^-1#

Explanation:

First we need to calculate the number of moles of #CO_2# in #1# tonne, which is #10^6# #g#. The molar mass of #CO_2# is #44# #gmol^-1#.

#n=m/M=10^6/44=22,727# #mol#

Now the total energy produced by #1# tonne is #1.68xx10^4# #MJ# = #1.68xx10^7# #kJ#. This is energy emitted to the environment, and therefore lost from the chemical system. That means it will take a negative value when we calculate the #DeltaH#.

We need to divide this by the number of moles to find the #DeltaH# value in #kJmol^-1#.

#DeltaH=-(1.68xx10^7)/(22,727)=-739# #kJmol^-1#