2019-04-03T10:35:51-04:00
A scientist measures the standard enthalpy change for the following reaction to be -112.0 kJ :
2NO(g) + O2(g) ------> 2NO2(g)
Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of NO2(g) is _________ kJ/mol.
1
2019-04-08T07:55:59-0400
Δ H f ( N O ) = 90.25 k J m o l \Delta H_f(NO)=90.25 \frac{kJ}{mol} Δ H f ( NO ) = 90.25 m o l k J
Δ H r x n = ∑ Δ H f 0 ( p r o d u c t s ) − ∑ Δ H f 0 ( r e a c t a n t s ) \Delta H_{rxn}= \sum \Delta H_f^0(products)-\sum \Delta H_f^0(reactants) Δ H r x n = ∑ Δ H f 0 ( p ro d u c t s ) − ∑ Δ H f 0 ( re a c t an t s ) Δ H r x n = 2 × Δ H f 0 ( N O 2 ) − ( 2 × Δ H f 0 ( N O ) + 1 × Δ H f 0 ( O 2 ) ) \Delta H_{rxn}= 2\times \Delta H_f^0(NO_2)-(2\times \Delta H_f^0(NO)+1\times \Delta H_f^0(O_2)) Δ H r x n = 2 × Δ H f 0 ( N O 2 ) − ( 2 × Δ H f 0 ( NO ) + 1 × Δ H f 0 ( O 2 ))
− 112.0 k J m o l = 2 × Δ H f 0 ( N O 2 ) − ( 2 × 90.25 k J m o l + 1 × 0 k J m o l ) -112.0 \frac{kJ}{mol}= 2\times\Delta H_f^0(NO_2)-(2\times 90.25 \frac{kJ}{mol}+1\times 0\frac{kJ}{mol}) − 112.0 m o l k J = 2 × Δ H f 0 ( N O 2 ) − ( 2 × 90.25 m o l k J + 1 × 0 m o l k J )
Δ H f 0 ( N O 2 ) = 34.25 k J m o l \Delta H_f^0(NO_2)= 34.25 \frac{kJ}{mol} Δ H f 0 ( N O 2 ) = 34.25 m o l k J
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