FOXBOROUGH, Mass. - Rob Gronkowski has been named AFC Offensive Player of the Week for his performance in the Patriots 27-24 victory at Pittsburgh on Sunday. It marks the second time in his career that he has been named AFC Offensive Player of the Week. He first earned the award in Week 14 of the 2011 season.
Gronkowski finished the game with nine receptions for a career-high 168 yards. The 168-yards pushed his 2017 total to 1,017 yards, giving him his fourth 1,000-yard season. Gronkowski joins Tony Gonzalez and Jason Witten as the only NFL tight ends with four 1,000-yard seasons. In addition, to his 168-yard performance, Gronkowski converted a key 2-point play after the game-winning touchdown to give the Patriots a 27-24 lead. During the game-winning drive, Gronkowski caught three passes for 69 yards with back-to-back 26-yard receptions followed by a 17-yard catch. He had four receptions of at least 20 yards in the game.
It is the sixth time in 2017 that the Patriots have been honored with a Player of the Week Award. QB Tom Brady was named AFC Offensive Player of the Week in Week 2, 3 and 10. RB Dion Lewis was named AFC Special Teams Player of the Week in Week 10, and K Stephen Gostkowski earned the award after his performance in Week 11. In addition, Brady was named AFC Offensive Player of the Month for November.