1. The enthalpy of the reaction 2 CH3OH(l) + 3O2(g) arrow 2 C02(g) + 4 H2O(l) is
ID: 907927 • Letter: 1
Question
1. The enthalpy of the reaction 2 CH3OH(l) + 3O2(g) arrow 2 C02(g) + 4 H2O(l) is -1452.8 k). A quantity of 453.6 g of liquid methanol is burned under constant pressure conditions. The amount of heat absorbed by the system in this process is: a. -658990 kJ b. -329495 kJ C. -20566 kJ d.-10283 kJ e. -5142 kj f. -1452.8 kj g. -726.4 kj h. +20566 kJ 2. The heat capacity of aluminum is 24.2 Jmol^-1K^-1. The temperature of a block of aluminum weighing 28.3 5 g decreases from 26.6 C to 3.3 C. The amount of heat absorbed by the aluminum block in this process is: a. -15985 J b. -686 J c.-676J d. -661 J e. -592 J f. -564 J g. .537 J h. 564 J 3. The following standard enthalpies of reaction have been obtained under standard conditions: 1. N2H4(g) + H2(g) arrow 2 NH3(g) deltaH 1= -187.6 kj 2. 3 112(g) + N2(g) arrow 2 NH1(g) deltaH2= -92.2 kj The following reaction is investigated 3. 2 H2(g) + N2(g) arrow N2H2(g) deltaH3 = ? The value of deltaH 3 is: a. -279.8kJ b. -95.4 kj c.47.7 kj d.95.4kJ e. 156.9 kj f. 218.3 kj g. 279.8 kj h. 470.6 kjExplanation / Answer
1) mol CH3OH = 453.6g /32g/mol= 14.175 mol
2 mol CH3OH --------- -1452.8 kJ
14.175 mol CH3OH --------- x= -10297 kJ (if you take the exact atomic mass from peridic table the result is -10283 kJ, i have rounded, so the correct option is d)
2) Heat absorbed= 28.35g x24.2 J/molK x -23.3K / 26.9815g/mol = -592 J
3) 2NH3 -----> H2 + N2H4 DH= 187.6 kJ
3H2 + N2 -----> 2NH3 DH= -92.2 kJ
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2H2 + N2 -------> N2H4 DH= 187.6 - 92.2= 95.4 kJ
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