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Calculate ΔrH for the following reaction: CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g) Use the following reactions and given ΔrH's. C(s)+2H2(g)→CH4(g)ΔrH=−74.6kJmol−1C(s)+2Cl2(g)→CCl4(g)ΔrH=−95.7kJmol−1H2(g)+Cl2(g)→2HCl(g)ΔrH=−92.3kJmol−1

Calculate ΔrH for the following reaction:

CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g)

Use the following reactions and given ΔrH's.

C(s)+2H2(g)→CH4(g)ΔrH=−74.6kJmol−1C(s)+2Cl2(g)→CCl4(g)ΔrH=−95.7kJmol−1H2(g)+Cl2(g)→2HCl(g)ΔrH=−92.3kJmol−1

Solutions

Expert Solution

C(s)+2H2(g)→CH4(g)ΔrH=−74.6kJmol−1 -------------> 1

C(s)+2Cl2(g)→CCl4(g)ΔrH=−95.7kJmol−1 ----------------->2

H2(g)+Cl2(g)→2HCl(g)ΔrH=−92.3kJmol−1 ------------------->3

3rd equation is multifly with 2 and than adding 2nd and 3rd equation

C(s)+2Cl2(g)----------->CCl4(g)                        ΔrH=−95.7kJmol−1 ----------------->2

2H2(g)+2Cl2(g)----------->4HCl(g)                ΔrH=−184.6kJmol−1 ------------------->3

--------------------------------------------------------------------------------------

C(s) + 2H2(g) + 4Cl2(g) -------->. CCl4(g) + 4HCl(g)       ΔrH   = -280.3KJ -----------> 4

C(s)+2H2(g)---------------->CH4(g)                                      ΔrH=−74.6kJmol−1 -------------> 1 substact

(-)      (-)                            (-)                                                     (+)

--------------------------------------------------------------------------------------------------------------------------------

CH4(g) + 4Cl2(g) -------------> CCl4(g) + 4HCl(g)                  ΔrH   = -205.7KJ mol^-1

ΔrH   = -205.7KJ mol^-1 >>>>>answer

queen_honey_blossom answered 2 years ago
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