Question

When 5.000 grams of ammonia react with an excess of oxygen and CH4 in a bomb calorimeter with a total heat capacity of 15.48 kJ/°C, the temperature of the calorimeter and its contents changes from 19.77°C to 33.90°C. Calculate ∆E and ∆H for the following reaction as written: 2 NH3 (g) + 3 O2 (g) + 2 CH4 (g) → 2 HCN (g) + 6 H2O (l) ∆E = ? ∆H = ? 2.

Given the following reaction: Zn (s) + 2 HBr (aq) → ZnBr2 (aq) + H2 (g) When 0.7240 grams of zinc are mixed with 75.00 mL of 0.8000 M HBr (aq) in a Styrofoam cup, the temperature of the solution inc

Answer 1

calculate

dE = C*dT

C = 15.48 kJ/°C

dT = (33.9°C - 19.77°C) = 14.13°C

dE = 15.48*14.13 = 218.7324 kJ

Now, calculade dH

dH = Q/n

m = 5 g of NH3

mol of NH3 = mass/MW = 5/17 = 0.29411 mol of NH3

dH = Q/n = 218.7324 / 0.29411 = 743.70949 kJ /mol of NH3

if written as 2 mol of NH3

then..

HRxn = 743.70949 kJ pe 2 mol of NHJ3

HRxn = 1487.41898 kJ per 2 mol of NH3