Suppose 0.562 grams of graphite is combusted in a calorimeter at 25 degreee C an

Question

Suppose 0.562 grams of graphite is combusted in a calorimeter at 25 degreee C and 0.5 atm pressure. The temperature of the calorimeter rises from 25 degreee C to 27.5 degreee C. If the heat capacity of the calorimeter is 20.7 J/degreee C, what is the heat of reaction? Express sour answer as a thermochemical equation (ie., write the balanced chemical equation). If 42.6 grams of Zn reacts in 75.0 ml of a 1.75 M solution of HCl at 25 degreee C and 0.852 atm according to the following equation: Zn (s) + 2 HCl rightarrow ZnCl_2(aq) + H_2 (g); Delta H = -153.7 kJ/mol Calculate q, w and Delta U for the reaction (1 atm = 101.325 KPa; 1 kJ = 1 kP * L; energy units are Joules).

Explanation / Answer

1. c(S) + o2   ----> CO2(g)

q = cp*DT

   = 20.7*(27.5-25) = 51.75 joule

no of mol of C = 0.562/12 = 0.0468 mol

DHrxn = 51.75*10^-3/0.0468 = 1.1 kj/mol

2. from the given data,

No of mol of Zn = 42.6/65.38 = 0.6515 mol

No of mol of HCl = 75/1000*1.75 = 0.13125 mol

limiting reactant = HCl

No of mol of ZnCl2 produced = 0.13125/2 = 0.0656 mol

q = 0.0656*153.7 = 10.08 kj

No of mol of H2 GAS PRODUCED = 0.0656 mol

volume of H2gas = nRT/P = (0.0656*0.0821*298)/0.852 = 1.88 l

DH = DU+PDV

0.852 atm = 86.3289 kpa

10.08 = X+86.3289*1.88

X = DU = -152.22 Kj

W = -PDV = -86.3289*1.88 = -162.3 Kj

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