New England Patriots tight end Rob Gronkowski was named the AFC Offensive Player of the Week for the second time in his career, the NFL announced on Wednesday.
Gronkowski reeled in nine receptions for a career-high 168 yards in the Patriots' 27-24 win over the Pittsburgh Steelers on Sunday. The 28-year-old has 64 catches for 1,017 yards and seven touchdowns in 12 games this season.