Question

A chemist measures the enthalpy change

ΔH

during the following reaction:

2Na (s) +   Cl2 (g) → 2NaCl (s)

=ΔH−822.kJ

Use this information to complete the table below. Round each of your answers to the nearest

/kJmol

.

A chemist measures the enthalpy change AH during the following reaction: 2Na(s) + Cl2(g)→2 NaCl(s) NH=-822. kJ Use this information to complete the table below. Round each of your answers to the nearest kJ/mol reaction △H kJ NaCl(s) → Na(s) + Cl2(g) kJ 2NaCl(s) → 2Na(s) + Cl2(g) kJ

Answer 1

given reaction is
2Na + Cl2 --> 2NaCl delta H = -822 KJ

1)
1/2Na + 1/4Cl2 --> 1/2NaCl
divide the coficiant of given reaction by 4 we will get this reaction
so, delta H of this reaction = (-822)/4
= -205.5 KJ

2)
NaCl --> Na + 1/2Cl2
first reverse the given reaction and then divide by 2
so, delta H of required of reaction is first multiplied by (-) and then divide it by 2
so, delta H of this reaction = -(-822)/2
= 411 KJ

3)
2NaCl --> 2Na + Cl2
just reverse the given reaction we will get this reaction
so, delta H of this reaction is negative of that of given reaction
so, delta H of this reaction = -(-822)
= 822 KJ