Be sure to answer all parts. consider the following balanced thermochemical equation for the decomposition of the mineral magne mgco3(s) → mgo(s) + co2(g) δhrxn = 117.3 kj (a) is heat absorbed or released in the reaction? released absorbed (b) what is δhrxn for the reverse reaction? kj (c) what is δh when 5.10 mol of co2 reacts with excess mgo? kj (d) what is δh when 41.0 g of co2 reacts with excess mgo?

a)the system absorbed heat

b)ΔHrxn = -117.3 kJ

c)ΔH =  603.33 kJ

d)ΔH= 109.3 kJ

Explanation:

Step 1: The balanced equation

MgCO3(s) → MgO(s) + CO2(g)

Step 2: Data given

ΔHrxn = 117.3 kJ

(a) Is heat absorbed or released in the reaction?

When ΔHrxn > 0 means the system absorbed heat. This is called an endothermic reaction.

b)  What is ΔHrxn for the reverse reaction?

ΔHrxn = -117.3 kJ, The reversed reaction is exothermic

(c) What is ΔH when 5.10 mol of CO2 reacts with excess MgO?

When 1 mol of CO2 reacts   ΔH= 117.3 kJ

When 5.10 mo of CO2 reacts   ΔH = 117.3 * 5.10 = 603.33 kJ

(d) What is ΔH when 41.0 g of CO2 reacts with excess MgO?

Moles = mass / Molar mass

Moles = 41.0 grams / 44.01 g/mol = 0.932 moles

For 1 mol  ΔH= 117.3 kJ

For 0.932 moles    ΔH= 117.3 * 0.932 = 109.3 kJ

what are the choices?

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years