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Compute the heat of reaction at 1000^(@) C foir 1/2H_(2)(g) + 1/2 Cl_(2)(g) to HCl(g), DeltaH_(298 K)^(@) = 92.236 kJ mol^(-1) C_(P)^(@)(H_(2),g) = 29.0284 - 0.8355 xx 10^(-3)T + 2.0097 xx 10^(-8) T^(2) C_(P)^(@) (Cl_(2),g) = 31.6555 + 1.0134 xx 10^(-2) T - 4.0337 xx 10^(-8) T^(2) C_(P)^(@)(HCl, g) = 28.1359 + 1.8078 xx 10^(-3) T + 1.5453 xx 10^(-8) T^(2) |
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Answer» <P> Solution :`DeltaH_(1273 K)^(@) = DeltaH_(298 K)^(@) + int_(298 K)^(1273 K) {(DeltaC_(p))DT}`where, `DeltaCO_(P) = C_(P)^(@)(HCl,g) -1/2C_(P)^(@)(H_(2),g) -1/2C_(P)^(@)(Cl_(2),g)` `=(28.1359 - 1/2 xx 29.0284 - 1/2 xx 31.6556) + (1.8078 + 1/2 xx 0.8355 - 1/2 xx 10.134) xx 10^(-3) T + (1.5453 - 1/2 xx 2.0097 + 1/2 xx 4.0337) xx 10^(-6) T^(2)` Hence, `DeltaH_(1273 K)^(@) =[-92.236 xx 10^(2) + (-2.2061) -(2.8415 xx 10^(-3))(1273^(3)/3 - 298^(3)/3)] J"mol"^(-1)` `=-92236 - 2151.0 + 2176.2 + 1736.0 = -94827 J "mol"^(-1) = -94.827 kJ "mol"^(-1)` |
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