Question

For the reaction C2H4+3O2=2CO2+2H2O, deltaE=-1415kJ.The deltaH at 27°C would be?

Answer 1

Answer : The enthalpy change is -1415 kJ

Explanation :

The given balanced chemical reaction is,

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)$

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = ?

$\Delta U$ = change in internal energy = $-1415kJ=-1415000J$

$\Delta n_g$ = change in moles = [(2+2)-(3+1)] = 0 (from reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27^oC=273+27=300K$

Now put all the given values in the above formula, we get:

$\Delta H=\Delta U+\Delta n_gRT$

$\Delta H=(-1415000J)-[0mol\times 8.314J/mol.K\times 373K$

$\Delta H=-1415000J$

$\Delta H=-1415kJ$

Therefore, the enthalpy change is -1415 kJ

Answer 2

Explanation:

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