What is the value for AG at 100 Kif AH = 27 kJ/mol and AS = 0.09 kJ/(mol-K)? A. AG = 2700 kJ/mol
B. AG = -36 kJ/mol
C. AG = 0 kJ/mol
D. AG = 18 kJ/mol

ΔG =   18KJ/mol

Explanation:

Given data:

ΔG = ?

ΔS = 0.09 Kj/mol.K

ΔH = 27 KJ/mol

Temperature  = 100 K

Solution:

Formula:

ΔG = ΔH - TΔS

ΔS = entropy

ΔH = enthalpy

by putting values,

ΔG =  27 KJ/mol - 100K(0.09 Kj/mol.K)

ΔG =  27 KJ/mol - 9 KJ/mol

ΔG =   18KJ/mol

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

the answer is c.the valence of the atoms combine .