Question
During complete combustion of one mole of butane, $$2658\,kJ$$ of heat is released. The thermochemical reaction for above change is
A.
$$2{C_4}{H_{10\left( g \right)}} + 13{O_{2\left( g \right)}} \to $$ $$8C{O_{2\left( g \right)}} + 10{H_2}{O_{\left( l \right)}};$$ $${\Delta _c}H = - 2658.0\,kJ\,mo{l^{ - 1}}$$
B.
$${C_4}{H_{10\left( g \right)}} + \frac{{13}}{2}{O_{2\left( g \right)}} \to $$ $$4C{O_{2\left( g \right)}} + 5{H_2}{O_{\left( g \right)}};$$ $${\Delta _c}H = - 1329.0\,kJ\,mo{l^{ - 1}}$$
C.
$${C_4}{H_{10\left( g \right)}} + \frac{{13}}{2}{O_{2\left( g \right)}}$$ $$ \to 4C{O_{2\left( g \right)}} + 5{H_2}{O_{\left( l \right)}};$$ $${\Delta _c}H = - 2658.0\,kJ\,mo{l^{ - 1}}$$
D.
$${C_4}{H_{10\left( g \right)}} + \frac{{13}}{2}{O_{2\left( g \right)}} \to $$ $$4C{O_{2\left( g \right)}} + 5{H_2}{O_{\left( l \right)}};$$ $${\Delta _c}H = + 2658.0\,kJ\,mo{l^{ - 1}}$$
Answer :
$${C_4}{H_{10\left( g \right)}} + \frac{{13}}{2}{O_{2\left( g \right)}}$$ $$ \to 4C{O_{2\left( g \right)}} + 5{H_2}{O_{\left( l \right)}};$$ $${\Delta _c}H = - 2658.0\,kJ\,mo{l^{ - 1}}$$
Solution :
$${\text{The}}\,\,{\text{value of}}\,\,{\Delta _c}H\,\,{\text{is}}\,\, - 2658\,kJ/mol.$$