A) 13 kJ
B) 16 kJ
C) 62 kJ
D) 21 kJ
Correct Answer: C
Solution :
\[{{C}_{p}}=23+0.01\,T\] |
\[n=3\]moles |
\[{{T}_{1}}=300K\] |
\[{{T}_{2}}=1000K\] |
\[\Delta H=n{{C}_{p}}\Delta T\] |
\[\Delta H=n\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{{{C}_{p}}}\Delta T=3\int\limits_{300}^{1000}{(23+0.01T)dT}\]\[=3\int\limits_{300}^{100}{\left( 23T+\frac{0.01{{T}^{2}}}{2} \right)}\]= 61.95 KJ/mle = 62 KJ/mole. |
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