Acetylene burns in air according to the following equation: c2h2(g) + 5 2 o2(g) → 2 co2(g) + h2o(g) δh o rxn = −1255.8 kj given

Acetylene burns in air according to the following equation: c2h2(g) + 5 2 o2(g) → 2 co2(g) + h2o(g) δh o rxn = −1255.8 kj given δh o f of co2(g) = −393.5 kj/mol and δh o f of h2o(g) = −241.8 kj/mol, find δh o f of c2h2(g).

Answers

 -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+frac{5}{2}O_2(g)rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

Delta H=sum [ntimes Delta H_f(product)]-sum [ntimes Delta H_f(reactant)]

Delta H=[(n_{CO_2}times Delta H_{CO_2})+ n_{H_2O}times Delta H_{H_2O})]-[(n_{C_2H_2}times Delta H_{C_2H_2})+(n_{O_2}times Delta H_{O_2})]

where,

n = number of moles

Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2times -393.5)+(1times -241.8)]-[(1times Delta H_{C_2H_2})+(frac{5}{2}times 0)]

-1255.8=[(-787)+(-241.8)]-[(1times Delta H_{C_2H_2})+(0)]

Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

its going to be mutualism cuz they both benefit from the relationship.

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you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

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